CCPC[长春] 2.Fraction 手动模拟

Fraction

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0


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

 

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直接手动模拟，上代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define PI acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
int dir[4][2]={{1,0},{-1,0},{0,1},{0,-1}};

ll a[100];
ll b[100];
int main()
{
int T,n,k,i,j;
scanf("%d",&T);
for(k=1;k<=T;k++)
{
memset(a,0,sizeof(a));
memset(b,0,sizeof(b));
ll p,q,pt,qt;
scanf("%d",&n);
for(i=1;i<=n;i++)
scanf("%lld",&a[i]);
for(i=1;i<=n;i++)
scanf("%lld",&b[i]);
ll t;

p=b
;
q=a
;

for(i=n-1;i>=1;i--)
{
p += q*a[i];

pt=p;
qt=q;

q=pt;
p=qt*b[i];
}

t = gcd(p,q);
p /= t;
q /= t;
printf("Case #%d: %lld %lld\n",k,p,q);
}

return 0;
}