新思维：C语言程序设计

5
1 2 3 4 5
1 2 3 4 5
9 2 2 4 5
1 2 7 4 5
1 5 6 4 8

#include <stdio.h>
int main(int argc, char *argv[])
{
    int a[3][3], sum_a, sum_b;
 
    scanf("%d%d%d", &a[0][0], &a[0][1], &a[0][2]);
    scanf("%d%d%d", &a[1][0], &a[1][1], &a[1][2]);
    scanf("%d%d%d", &a[2][0], &a[2][1], &a[2][2]);
 
    sum_a = a[0][0]+a[1][1]+a[2][2];
    sum_b = a[0][2]+a[1][1]+a[2][0];
 
    printf("%d %d\n", sum_a, sum_b);
 
    return 0;
}

#include <stdio.h>
 
#define N  100
int main(int argc, char *argv[])
{
    int a[N][N];
    int n, i, j, sum_a=0, sum_b=0;
 
    scanf("%d", &n);
    for (i=0; i<n; i++)
        for (j=0; j<n; j++)
            scanf("%d", &a[i][j]);
 
    for (i=0; i<n; i++) {
        sum_a = sum_a + a[i][i];
        sum_b = sum_b + a[i][n-1-i];        
    }     
 
    printf("%d %d\n", sum_a, sum_b);
 
    return 0;
}

int main(int argc, char *argv[])
{
    int v[6] = { 2, 3, 9, 1, 0, 7};
    int i, j, n, t;
 
    for (j=0; j<5; j++)
        if (v[j]>v[j+1])
            t = v[j], v[j] = v[j+1], v[j+1]=t;
 
    for (i=0; i<6; i++)
        printf("%d ", v[i]);
    printf("\n");
    return 0;
}