循环小数

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循环小数
Time Limit : 3000/1000ms (Java/Other) Memory Limit : 65535/32768K (Java/Other)
Total Submission(s) : 15 Accepted Submission(s) : 4

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Problem Description

Given a floating point number X, the form is X=0.a1a2...an(b1b2...bm)(1<=n,m<=8). (b1b2...bm) indicates the repetend. For example, 0.5=0.50=0.5(0)=0.5(00)=1/2,0.3(3)=0.333(33)=1/3.
Please change the floating point number X into a fraction A/B(gcd(A,B)=1).

Input

Input may contain several test cases. The first line is a positive integer, T(T<=10) indicating the number of test cases below.
For each test case, the first line contains two integers n,m(1<=n,m<=8).
The following line contains n numbers a1a2...an.
The following line contains m numbers b1b2...bm.

Output

For each case, output the number A and B(0<A,B) on a single line.

Sample Input

22 250004 2333333

Sample Output

1 21 3

Author

//无限循环小数化为分数

参考这里: http://wenku.baidu.com/view/561773c62cc58bd63186bd87.html

#include<cstdio>__int64 solve(__int64 a,__int64 b){    return b==0?a:solve(b,a%b);}int main(){    int t,n,m;    __int64 a,b,z,x,y;    scanf("%d",&t);    while(t--)    {        scanf("%d%d%I64d%I64d",&n,&m,&a,&b);        x=a;        y=0;        while(m--)        {            x*=10;            y=y*10+9;        }        x=x+b-a;        while(n--)        {            y*=10;        }        z=solve(x,y);        //printf("%I64d %I64d\n",x,y);        printf("%I64d %I64d\n",x/z,y/z);    }    return 0;}


 

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