hdu 5912 Fraction 模拟题

Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1045    Accepted Submission(s): 468

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

121 12 3

 

Sample Output

Case #1: 1 2
Hint
Here are the details for the first sample:2/(1+3/1) = 1/2

      签到题

#include <iostream>#include <cstdio>using namespace std;long long int a[15];long long int b[15];long long int gcd(long long int a,long long int b){    return b==0?a:gcd(b,a%b);}int main(){    int t;    scanf("%d",&t);    int ca=1;    while(t--)    {        int n;        scanf("%d",&n);        for(int i=1; i<=n; i++)        {            scanf("%I64d",&a[i]);        }        for(int i=1; i<=n; i++)        {            scanf("%I64d",&b[i]);        }        printf("Case #%d:",ca++);        long long int son,mom;        son=b[n];        mom=a[n];        int cnt=n;        long long int x=1;        while(cnt>1)        {            x=gcd(son,mom);            son/=x;            mom/=x;            son+=(mom*a[cnt-1]);            long long int ss=son;            son=mom*b[cnt-1];            mom=ss;            cnt--;        }        x=gcd(son,mom);        son/=x;        mom/=x;        printf(" %I64d %I64d\n",son,mom);    }    return 0;}