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  • Codeforces Round #395 (Div. 2)

    题目链接:http://codeforces.com/contest/764/problem/A

    题意:有两个人,第一个人每n分钟到达一次目的地,第二个人每m分钟到达一次目的地,现在给定一个时间z问在z这个时间范围内两人相遇的次数。

    思路:由于z,n,m的范围比较小。所以直接开个数组暴力模拟就好了。  更简单的其实答案就是z/lcm(n,m)

    #define _CRT_SECURE_NO_DEPRECATE
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<stdio.h>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<map>
    #include<set>
    #include<time.h>
    #include<cmath>
    using namespace std;
    #define x first
    #define y second
    #define pb push_back
    #define mp make_pair
    typedef long long int LL;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3fLL;
    const int MAXN = 1e4 + 10;
    int v[MAXN];
    int main(){
    //#ifdef kirito
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt", "w", stdout);
    //#endif
    //    int start = clock();
        int n,m,z;
        while (~scanf("%d%d%d", &n,&m,&z)){
            memset(v, 0, sizeof(v));
            for (int i = 1; i*n <= z; i++){
                v[i*n]++;
            }
            int ans = 0;
            for (int i = 1; i*m <= z; i++){
                if (v[i*m] > 0){
                    ans++;
                }
            }
            printf("%d
    ", ans);
        }
    //#ifdef LOCAL_TIME
    //    cout << "[Finished in " << clock() - start << " ms]" << endl;
    //#endif
        return 0;
    }
    #define _CRT_SECURE_NO_DEPRECATE
    #include<iostream>
    #include<cstring>
    #include<string>
    #include<algorithm>
    #include<stdio.h>
    #include<queue>
    #include<vector>
    #include<stack>
    #include<map>
    #include<set>
    #include<time.h>
    #include<cmath>
    using namespace std;
    #define x first
    #define y second
    #define pb push_back
    #define mp make_pair
    typedef long long int LL;
    const int inf = 0x3f3f3f3f;
    const LL INF = 0x3f3f3f3f3f3f3f3fLL;
    const int MAXN = 1e5 + 10;
    int gcd(int x, int y){
        return y==0? x : gcd(y, x%y);
    }
    int lcm(int x, int y){
        return x*y / gcd(x, y);
    }
    int main(){
    //#ifdef kirito
    //    freopen("in.txt", "r", stdin);
    //    freopen("out.txt", "w", stdout);
    //#endif
    //    int start = clock();
        int n,m,z;
        while (~scanf("%d%d%d", &n,&m,&z)){
            printf("%d
    ", z / lcm(n, m));
        }
    //#ifdef LOCAL_TIME
    //    cout << "[Finished in " << clock() - start << " ms]" << endl;
    //#endif
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/kirito520/p/6363481.html
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