110120
UID: 3431, 注册于 2021-10-31 22:12:34, 最后登录于 2023-6-26 16:28:29, 目前离线.
解决了 1 道题目，RP: 11 (No. 2230)
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个人简介
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MODelse: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MODelse: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )v
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )
MOD = 1000000007
小于等于n的所有数字的平方和
def get_sum(n): if n == 0: return 0, 0, 0
arr = [] val = n while val: arr.append(val % 10) val //= 10
前i个位置做选择，选择出的数字要求数位总和模7不等于bit_sum_mod7
且数字本身模7不等于val_sum_mod7, 在所有高位都选择了最大值的标记为all_higher_max
的情况下，所有合法数字的平方和，平方和是可以拆解成每个子集的数的平方和来求解的
函数需要返回集合中所有数的数量，一次项累加和以及二次项累加和
from functools import lru_cache @lru_cache(typed=False, maxsize=128000000) def dp(i, bit_sum_mod7, val_sum_mod7, all_higher_max): if i == 0: upper_bound = arr[i] if all_higher_max else 9
cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound+1): if cur_val == 7: continue if cur_val % 7 == bit_sum_mod7: continue if cur_val % 7 == val_sum_mod7: continue cnt += 1 sum += cur_val square_sum += cur_val * cur_val return cnt, sum % MOD, square_sum % MOD else: upper_bound = arr[i] if all_higher_max else 9 cnt = 0 sum = 0 square_sum = 0 for cur_val in range(upper_bound + 1): if cur_val == 7: continue new_cnt, new_sum, new_square_sum = dp(i-1, (bit_sum_mod7 - cur_val) % 7, (val_sum_mod7 - cur_val * 10 ** (i)) % 7, all_higher_max and cur_val == arr[i]) cnt += new_cnt sum += new_cnt*(cur_val * (10**(i))) + new_sum square_sum += new_cnt * ( (cur_val**2) * (10**(2*i)) ) + 2*cur_val*(10**(i))*new_sum + new_square_sum return cnt, sum % MOD, square_sum % MODreturn dp(len(arr)-1, 0, 0, True) T = int(input()) for _ in range(T): a, b = map(int, input().split()) print( (get_sum(b)[2] - get_sum(a-1)[2]) % MOD )v
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