Gym - 100735E Restore

Posted 2020-08-31 gggyt

E - Restore

题意：输入一个n，输入一个对角线空缺（为0）的n*n的矩阵，要求每一行每一列和对角线的和相同，输出完整的矩阵。

解法：设每一行的和都是sum，用一个h[]数组存每一行的和。则可得a[0][0] = sum-h[0], a[1][1] = sum-h[1], a[2][2] = sum-h[2]......同时所有对角线的和也为sum，则可得公式 sum-h[0]+sum-h[1]+...+sum-h[n-1] = sum, 设所有h[x]的和为summ, 即n*sum-summ=sum, 即可得sum = summ/(n-1)，后面由式子a[i][i] = sum-h[i]便可求得。

注意：a[i][j]的取值范围！?-?10¹²?≤?A_ij?≤?10^{12， 要用long long去存。}

 

typedef long long ll;
ll a[105][105];
ll h[105];
void solve() {
    int n;  scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            scanf("%lld", &a[i][j]);
            h[i] += a[i][j];
        }
    }
    ll summ=0;
    for (int i = 0; i < n; i++) {
        summ+=h[i];
    }
    ll sum = summ/(n-1);
    for (int i = 0; i < n; i++) {
        a[i][i] = sum-h[i];
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (j)  printf(" ");
            printf("%lld", a[i][j]);
        }
        printf("\n");
    }
}
int main() {
    int t = 1;
    //scanf("%d", &t);
    while(t--)
        solve();
    return 0;
}