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; }