Wannafly挑战赛18 C - 异或和

思路：我刚开始是想旋转四次坐标，每次用bit计算每个点左上角的点到这个点的距离，TLE了。。。。

这种算曼哈顿距离的可以将x 轴和 y 轴独立开来，分别计算。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 2000 + 7;
const int M = 1e6 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +7;

int n, m;
LL x[N], y[N], sumx[N], fsumx[N], sumy[N], fsumy[N], num;
char s[N][N];

inline void add(LL &a, LL b) {
    a += b;  if(a >= mod) a -= mod;
}
LL fastPow(LL a, LL b) {
    LL ans = 1;
    while(b) {
        if(b & 1) ans = ans * a % mod;
        a = a * a % mod; b >>= 1;
    }
    return ans;
}
int main(){
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) {
        scanf("%s", s[i] + 1);
    }

    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            if(s[i][j] == '1') {
                x[i]++;
                y[j]++;
                num++;
            }
        }
    }

    LL cnt = 0;
    for(int i = 1; i <= n; i++) add(sumx[i], sumx[i - 1]), add(sumx[i], cnt), add(cnt, x[i]);

    cnt = 0;
    for(int i = n; i >= 1; i--) add(fsumx[i], fsumx[i + 1]), add(fsumx[i], cnt), add(cnt, x[i]);

    cnt = 0;
    for(int i = 1; i <= m; i++) add(sumy[i], sumy[i - 1]), add(sumy[i], cnt), add(cnt, y[i]);

    cnt = 0;
    for(int i = m; i >= 1; i--) add(fsumy[i], fsumy[i + 1]), add(fsumy[i], cnt), add(cnt, y[i]);

    num = fastPow(num, mod - 2);

    LL ans = 0;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            ans ^= (sumx[i] + fsumx[i] + sumy[j] + fsumy[j]) * num % mod;
        }
    }

    printf("%lld\n", ans);
}
/*
*/