100 Accepted
# | 状态 分数 | 耗时 | 内存占用 |
---|---|---|---|
#1 | Accepted 10 | 2ms | 536 KiB |
#2 | Accepted 10 | 3ms | 536 KiB |
#3 | Accepted 10 | 2ms | 536 KiB |
#4 | Accepted 10 | 2ms | 324 KiB |
#5 | Accepted 10 | 2ms | 764 KiB |
#6 | Accepted 10 | 2ms | 536 KiB |
#7 | Accepted 10 | 2ms | 324 KiB |
#8 | Accepted 10 | 3ms | 536 KiB |
#9 | Accepted 10 | 2ms | 536 KiB |
#10 | Accepted 10 | 3ms | 536 KiB |
代码
#include<iostream>
using namespace std;
const int N=1e2+3;
int a[N][N],h,k=1,l,hy=0,b=2,n;
int main(){
scanf("%d",&n);
h=n-1;
for(int i=0;i<=n-1;i++){
h=n-hy;
l=1;
for(;l!=b;){
a[h][l]=k++;
l++;
}
l--;
for(;h!=n;){
h++;
a[h][l]=k++;
}
b++;
hy++;
}
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++)
printf("%4d",a[i][j]);
printf("\n");
}
return 0;
}
信息
- 递交者
- 题目
- P369 练63.3 迂回方阵
- 语言
- C++14(O2)
- 代码长度
- 387 Bytes
- 递交时间
- 2024-11-22 21:01:57
- 评测时间
- 2024-11-22 21:02:11
- 分数
- 100
- 总耗时
- 24ms
- 峰值时间
- 3ms
- 峰值内存
- 764 KiB