zoukankan      html  css  js  c++  java
  • Educational Codeforces Round 45 (Rated for Div. 2) G GCD Counting

    G - GCD Counting

    思路:我猜测了一下gcd的个数不会很多,然后我就用dfs回溯的时候用map暴力合并就好啦。

    终判被卡了MLE。。。。。  需要每次清空一下子树的map。。。

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define pii pair<int,int>
    #define piii pair<int, pair<int,int> >
    
    using namespace std;
    
    const int N = 2e5 + 7;
    const int M = 10 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-6;
    
    int n, a[N];
    LL cnt[N];
    vector<int> edge[N];
    map<int, LL> mp[N];
    
    int gcd(int a, int b) {
        return !b ? a : gcd(b, a % b);
    }
    
    void dfs(int u, int p) {
        mp[u][a[u]]++;
        for(int v : edge[u]) {
            if(v == p) continue;
            dfs(v, u);
            map<int, LL> :: iterator itu, itv;
            for(itu = mp[u].begin(); itu != mp[u].end(); itu++) {
                for(itv = mp[v].begin(); itv != mp[v].end(); itv++) {
                    cnt[gcd(itu -> first, itv -> first)] += itu -> second * itv -> second;
                }
            }
    
            for(itv = mp[v].begin(); itv != mp[v].end(); itv++) {
                mp[u][gcd(a[u], itv -> first)] += itv -> second;
            }
        }
    }
    
    int main() {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            cnt[a[i]]++;
        }
    
        for(int i = 1; i < n; i++) {
            int u, v; scanf("%d%d", &u, &v);
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
    
        dfs(1, 0);
    
        for(int i = 1; i < N; i++) {
            if(cnt[i]) {
                printf("%d %lld\n", i, cnt[i]);
            }
        }
        return 0;
    }
    /*
    */

    通过代码

    #include<bits/stdc++.h>
    #define LL long long
    #define fi first
    #define se second
    #define mk make_pair
    #define pii pair<int,int>
    #define piii pair<int, pair<int,int> >
    
    using namespace std;
    
    const int N = 2e5 + 7;
    const int M = 10 + 7;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3f;
    const int mod = 1e9 + 7;
    const double eps = 1e-6;
    
    int n, a[N];
    LL cnt[N];
    vector<int> edge[N];
    map<int, LL> mp[N];
    
    int gcd(int a, int b) {
        return !b ? a : gcd(b, a % b);
    }
    
    void dfs(int u, int p) {
        mp[u][a[u]]++;
        for(int v : edge[u]) {
            if(v == p) continue;
            dfs(v, u);
            map<int, LL> :: iterator itu, itv;
            for(itu = mp[u].begin(); itu != mp[u].end(); itu++) {
                for(itv = mp[v].begin(); itv != mp[v].end(); itv++) {
                    cnt[gcd(itu -> first, itv -> first)] += itu -> second * itv -> second;
                }
            }
    
            for(itv = mp[v].begin(); itv != mp[v].end(); itv++) {
                mp[u][gcd(a[u], itv -> first)] += itv -> second;
            }
            mp[v].clear();
        }
    }
    
    int main() {
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            cnt[a[i]]++;
        }
    
        for(int i = 1; i < n; i++) {
            int u, v; scanf("%d%d", &u, &v);
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
    
        dfs(1, 0);
    
        for(int i = 1; i < N; i++) {
            if(cnt[i]) {
                printf("%d %lld\n", i, cnt[i]);
            }
        }
        return 0;
    }
    /*
    */
  • 相关阅读:
    【总结】数组去重的3种方式
    【原】运动版的轮播图,有左右按钮和单独分页,原生JS版
    【巩固】JS中的封闭空间
    IIS 配置.svc的MIME映射
    c# 线程定时器 System.Threading.Timer 转载
    QTcreator快捷操作,转载
    C#中的钩子说明
    anaconda更新库命令
    Chart控件,chart、Series、ChartArea曲线图绘制的重要属性介绍
    c# chart控件柱状图,改变柱子宽度
  • 原文地址:https://www.cnblogs.com/CJLHY/p/9166880.html
Copyright © 2011-2022 走看看