题目地址：点击打开链接

因为数据量很小，直接倒着推模拟即可。

代码：

#include<bits/stdc++.h> using namespace std; typedef long long ll; const int maxn = 15; ll a[maxn], b[maxn], n; int main(void) { int ca = 1, t; cin >> t; while(t--) { scanf("%lld", &n); for(int i = 1; i <= n; i++) scanf("%lld", &a[i]); for(int i = 1; i <= n; i++) scanf("%lld", &b[i]); ll fz = b[n], fm = a[n]; fz /= __gcd(a[n], b[n]); fm /= __gcd(a[n], b[n]); for(int i = n; i > 1; i--) { ll nfz = fm*b[i-1]; ll nfm = fm*a[i-1]+fz; fz = nfz/__gcd(nfz, nfm); fm = nfm/__gcd(nfz, nfm); } printf("Case #%d: %lld %lld\n", ca++, fz, fm); } return 0; }

Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 836    Accepted Submission(s): 457 Problem Description Mr. Frog recently studied how to add two fractions up, and he came up with an evil idea to trouble you by asking you to calculate the result of the formula below: As a talent, can you figure out the answer correctly?   Input The first line contains only one integer T, which indicates the number of test cases. For each test case, the first line contains only one integer n ( n≤8 ). The second line contains n integers:  a1,a2,⋯an(1≤ai≤10 ). The third line contains n integers:  b1,b2,⋯,bn(1≤bi≤10) .   Output For each case, print a line “Case #x: p q”, where x is the case number (starting from 1) and p/q indicates the answer. You should promise that p/q is irreducible.   Sample Input 1 2 1 1 2 3   Sample Output Case #1: 1 2 Hint Here are the details for the first sample: 2/(1+3/1) = 1/2   Source 2016中国大学生程序设计竞赛（长春）-重现赛