HDU 1713 相遇周期 (最小公倍数)
HDU 最小 周期 公倍数
2023-09-11 14:17:18 时间
题意:。。。
析:求周期就是这两个分数的最小公倍数,可以先通分,再计算分子的最小倍数。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <unordered_map> #include <unordered_set> #define debug() puts("++++"); #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 5; const int mod = 2000; const int dr[] = {-1, 1, 0, 0}; const int dc[] = {0, 0, 1, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int main(){ int T; cin >> T; while(T--){ LL a, b, c, d; scanf("%lld/%lld", &a, &b); scanf("%lld/%lld", &c, &d); LL tmp = b / __gcd(b, d) * d; a = a * (tmp / b); c = c * (tmp / d); LL t = a / __gcd(a, c) * c; LL x = __gcd(t, tmp); t /= x; tmp /= x; if(1 == tmp) cout << t << endl; else cout << t << "/" << tmp << endl; } return 0; }
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