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); } /* */