114 条题解
-
-1
20220747 @ 2023-3-3 18:49:59
这题很简单，输入 aa 和 bb，然后输出他们的和即可。
#include <stdio.h> int main() { int a,b; scanf("%d%d",&a,&b); printf("%d\n", a+b); return 0; } -
-1
fengxiaoyi-T&JOI
@ 2022-12-3 11:34:28
#include<bits/stdc++.h> using namespace std; int main(){ int a,b; scanf("%d%d",&a,&b); return !printf("%d",a+b); } -
-1
FinalEnd_0346
@ 2022-4-11 7:50:21
#include<bits/stdc++.h> using namespace std; int main(){ int a,b; cin>>a>>b; cout<<a+b; return 0; } -
-1
LAOFENG @ 2022-3-26 0:01:54
//新手可看此贴 #include<bits/stdc++.h> using namespace std; long long a,b;//定义变量a,b int main(){ cin>>a>>b;//将a,b输入 cout<<a+b;//将a,b的和输出 return 0;//主函数返回值为0 } -
-1
wujunhaoturing LV 2 @ 2021-10-23 14:47:13
#include<iostream> using namespace std; int main(){ int a,b; cin>>a>>b; cout<<a+b<<endl; return 0; } -
-1
_JY_ @ 2021-10-5 9:20:35
水个 py 的吧。
a=input().split() # 用 split() 切片 print(int(a[0])+int(a[1])) # 先转 int 型再相加 -
-1
SHM_15105854553 LV 4 @ 2021-4-19 6:47:31
import java.util.Scanner; public class Main { private static Scanner s; public static void main(String[] args) { s = new Scanner(System.in); int a = s.nextInt(); int b = s.nextInt(); System.out.println(a + b); } } -
-2
C++ 式普通写法
#include<iostream> using namespace std; int a,b; int main(){ cin >> a >> b; cout << a + b; return 0; }高精度
#include<iostream> using namespace std; const int MAXN = 1e5 + 10; long long r; int c; char op; struct bigint{ long long len,sz[MAXN] = {0}; long long &operator [] (long long a){ return sz[a]; } void input(){ string s; cin >> s; len = s.size(); for (int i = 0; i < len; i++){ sz[i] = s[len - i - 1] - '0'; } } void output(){ for (; len - 1 > 0 && !sz[len - 1]; len--){ } for (int i = len - 1; i >= 0; i--){ cout << sz[i]; } cout << '\n'; } bigint operator + (const bigint &b){ bigint c; c.len = max(len,b.len) + 1; for (int i = 0; i < c.len; i++){ c[i] = sz[i] + b.sz[i]; } for (int i = 0; i < c.len - 1; i++){ c[i + 1] += c[i] / 10,c[i] %= 10; } return c; } }a,b,ans; int main(){ a.input(),b.input(); ans = a + b; ans.output(); return 0; } -
-2
Wang Yize @ 2023-7-19 14:33:53
#include<bits/stdc++.h> using namespace std; int main(){ long long a , b ;//long long 的忠实粉丝 cin >> a >> b ;//cincout就是香！！ cout << a+b ; return 0; } -
-2
wangjun4hao LV 6 @ 2023-7-18 18:53:00
#include <bits/stdc++.h> using namespace std; int main() { //以下三行为输入输出超大数据时标准输入输出流的加速 ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); long long a,b; cin>>a>>b; cout<<a+b<<endl; return 0; } -
-2
**#include** **<**bits**/**stdc**++.**h**>** **using** **namespace** **std**; **/\*** ** **思路：将字符串s所有拆分的可能性都尝试一下 ** **打擂台求出最小的素数 ** **\*/ //素数判断 **bool** **sushu**(**int** **n**)** **{ ** **for** **(**int** **i** **=** **2**;** **i** **<=** **sqrt**(**n**);** **i**++)** **{ ** **if** **(**n** **%** **i** **==** **0**)** **{ ** **return** **false**;** ** **} ** **} ** **if** **(**n** **<=** **1**)** **return** **false**;** ** **return** **true**;** } **int** **main**()** **{ ** **string** **s**,** **s1**,** **s2**;** ** **int** **i**,** **mi** **=** **INT\_MAX**;** **//mi存储最小的素数 ** **cin** **>>** **s**;** ** **int** **x**,** **y**; ** **/\*长度为s.size()的字符串，拆s.size()-1次 ** **12345 ** **i=0** **1** **2345** **s.substr(0,1)** **s.substr(1) ** **i=1** **12** **345** **s.substr(0,2)** **s.substr(2) ** **\*/ ** **//循环拆段的次数 ** **for** **(**i** **=** **0**;** **i** **<** **s**.**size**()** **-** **1**;** **i**++)** **{** ** **s1** **=** **s**.**substr**(**0**,** **i** **+** **1**); ** **s2** **=** **s**.**substr**(**i** **+** **1**);** ** **//cout<<s1<<"** **"<<s2<<endl; ** **x** **=** **atoi**(**s1**.**c\_str**());** ** **y** **=** **atoi**(**s2**.**c\_str**());** ** **if** **(**sushu**(**x** **+** **y**)** **&&** **x** **+** **y** **<** **mi**)** **mi** **=** **x** **+** **y**; ** **} ** **if** **(**mi** **==** **INT\_MAX**)** **cout** **<<** **-**1**; ** **else** **cout** **<<** **mi**;** ** **return** **0**;** } -
-2
很简单，线段树模板题。
#include <bits/stdc++.h> #define int long long using namespace std; int x, y, a[100005]; struct node { int l, r, w, tag; } tr[400005]; inline bool inrange(int l1, int r1, int l2, int r2) { return (l1 >= l2) && (r1 <= r2); } inline bool outofrange(int l1, int r1, int l2, int r2) { return (l1 > r2) || (l2 > r1); } inline void pushup(int u) { tr[u].w = tr[u * 2].w + tr[u * 2 + 1].w; } inline void build(int u, int l, int r) { if (l == r) { tr[u].w = a[l]; return; } int mid = l + r >> 1; build(u * 2, l, mid); build(u * 2 + 1, mid + 1, r); pushup(u); } inline int query1(int u, int l, int r, int p) { if (l == r) { return tr[u].w; } else { int mid = l + r >> 1; if (p <= mid) return query1(u * 2, l, mid, p); else return query1(u * 2 + 1, mid + 1, r, p); } } inline void update1(int u, int l, int r, int p, int x) { if (l == r) { tr[u].w += x; } else { int mid = l + r >> 1; if (mid >= p) update1(u * 2, l, mid, p, x); else update1(u * 2 + 1, mid + 1, r, p, x); } } inline void maketag(int u, int len, int x) { tr[u].tag += x; tr[u].w += len * x; } inline void pushdown(int u, int l, int r) { int mid = l + r >> 1; maketag(u * 2, mid - l + 1, tr[u].tag); maketag(u * 2 + 1, r - mid, tr[u].tag); tr[u].tag = 0; } inline int query(int u, int l1, int r1, int l2, int r2) { if (inrange(l1, r1, l2, r2)) { return tr[u].w; } else if (!outofrange(l1, r1, l2, r2)) { int mid = l1 + r1 >> 1; pushdown(u, l1, r1); return query(u * 2, l1, mid, l2, r2) + query(u * 2 + 1, mid + 1, r1, l2, r2); } } inline void update(int u, int l1, int r1, int l2, int r2, int x) { if (inrange(l1, r1, l2, r2)) { maketag(u, r1 - l1 + 1, x); } else if (!outofrange(l1, r1, l2, r2)) { int mid = l1 + r1 >> 1; pushdown(u, l1, r1); update(u * 2, l1, mid, l2, r2, x); update(u * 2 + 1, mid + 1, r1, l2, r2, x); pushup(u); } } signed main() { cin >> x >> y; build(1, 1, 1); update1(1, 1, 1, 1, x); update1(1, 1, 1, 1, y); cout << query1(1, 1, 1, 1); return 0; }-
liyixuan922527 @ 2023-8-26 19:44:02
666
-
