1 条题解
-
0
C :
#include <stdio.h> #include <math.h> #include <stdlib.h> #include <time.h> #define ARR_SIZE 20 int main() { // freopen("in","r",stdin); // freopen("out","w",stdout); int a[ARR_SIZE][ARR_SIZE], i, j, n, sum = 0; long product = 1; while(scanf("%d", &n)!=EOF){ sum = 0;product = 1; for (i=0; i<n; i++) { for (j=0; j<n; j++) { scanf("%d",&a[i][j]); } } for (i=0; i<n; i++) { for (j=0; j<n; j++) { if (i == j || i+j == n-1) sum += a[i][j]; if ((i == j || i+j == n-1) && i%2 == 0 && j%2 == 0) product *= a[i][j]; } } printf("%d %ld\n", sum, product); } return 0; }
C++ :
#include<cstdio> using namespace std; int main() { int n; while (scanf("%d", &n) != EOF) { int arr[15][15]; int i, j; int sum = 0; int pro = 1; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { scanf("%d", &arr[i][j]); if (i + j == n - 1||i==j) { sum += arr[i][j]; if (i % 2 == 0 && j % 2 == 0) pro *= arr[i][j]; } } } printf("%d %d\n", sum, pro); } }
- 1
信息
- ID
- 257
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者