zoukankan      html  css  js  c++  java
  • Wannafly挑战赛18 C

    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
    ", ans);
    }
    /*
    */
  • 相关阅读:
    771. Jewels and Stones
    706. Design HashMap
    811. Subdomain Visit Count
    733. Flood Fill
    117. Populating Next Right Pointers in Each Node II
    250. Count Univalue Subtrees
    94. Binary Tree Inorder Traversal
    116. Populating Next Right Pointers in Each Node
    285. Inorder Successor in BST
    292. Nim Game Java Solutin
  • 原文地址:https://www.cnblogs.com/CJLHY/p/9217050.html
Copyright © 2011-2022 走看看