你可知“茴”字有几种写法

一道水题，但是在编码的时候居然卡住了，收集了一些写法，还满有趣的。

我自己的

vector

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 7;
const ll mod = 1e9 + 7;
inline ll read() {
    ll s = 0, f = 1;
    char ch;
    do {
        ch = getchar();
        if (ch == '-') f = -1;
    } while (ch < 48 || ch > 57);
    while (ch >= 48 && ch <= 57)
        s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return s * f;
}
vector<vector<ll>> a;
int main() {
    vector<ll> line;
    ll n = read(), m = read();
    for (int i = 0; i < n; ++i) {
        vector<ll> aa;
        ll x, sum = 0;
        for (int j = 0; j < m; ++j) {
            x = read();
            aa.push_back(x);
            sum += x;
        }
        line.push_back(sum);
        a.push_back(aa);
    }
    vector<ll> row;
    for (int j = 0; j < m; ++j) {
        ll sum = 0;
        for (int i = 0; i < n; ++i) sum += a[i][j];
        row.push_back(sum);
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) printf("%lld ", line[i] + row[j] - a[i][j]);
        printf("\n");
    }
    return 0;
}

YWG的

我觉得是最好的

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 7;
inline ll read() {
    ll s = 0, f = 1;
    char ch;
    do {
        ch = getchar();
        if (ch == '-') f = -1;
    } while (ch < 48 || ch > 57);
    while (ch >= 48 && ch <= 57)
        s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
    return s * f;
}
ll x[maxn], y[maxn], tmp;
int main() {
    ll n=read(), m=read();
    ll Map[n + 1][m + 1];
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= m; ++j) {
            Map[i][j]=read();
            x[i] += Map[i][j];
            y[j] += Map[i][j];
        }
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) printf("%lld ", x[i] + y[j] - Map[i][j]);
        printf("\n");
    }
    return 0;
}

青竹大佬的

#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 7;
typedef long long ll;
int n, m;
vector<int> v[1000000];
vector<ll> H, L;
int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        ll sum = 0;
        for (int j = 1, x; j <= m; j++) {
            scanf("%d", &x);
            v[i].push_back(x);
            sum += x;
        }
        H.push_back(sum);
    }
    for (int j = 1; j <= m; j++) {
        ll sum = 0;
        for (int i = 1; i <= n; i++) {
            sum += v[i][j - 1];
        }
        L.push_back(sum);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            ll ans = H[i - 1] + L[j - 1] - v[i][j - 1];
            printf("%lld ", ans);
        }
        printf("\n");
    }
}

出题人的

知识点：预处理。
思路：开两个数组分别处理每一行的和以及每一列的和。之后对应顶点输出（对应行+对应列-该顶点）即可。
这道题更难的地方是n和m大小未定。解决方法有两种，一个是动态内存（malloc或者vector），另一种方法是用一维数组代替二维数组（a[i][j]和a[i*m+j]等价）。这里标程用的是第二种方法。
标程代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
ll a[1000010];
ll sr[1000010],sc[1000010];
int main(){
    int n,m,i;
    cin>>n>>m;
    for(i=0;i<n*m;i++)scanf("%lld",&a[i]);
    for(i=0;i<n*m;i++){
        sr[i/m]+=a[i];
        sc[i%m]+=a[i];
    }
    for(i=0;i<n*m;i++){
        printf("%lld ",sr[i/m]+sc[i%m]-a[i]);
        if(i%m==m-1)printf("\n");
    }
}