循环小数

Problem DescriptionGiven 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).InputInput 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.OutputFor each case, output the number A and B(0<A,B) on a single line.Sample Input22 250004 2333333Sample Output1 21 3/*题解：一个混循环小数的小数部分可以化成分数，这个分数的分子是第二个循环节以前的小数部分组成的数与小数部分中不循环部分组成的数的差。分母的头几位数是9，末几位是0。9的个数与循环节中的位数相同，0的个数与不循环部分的位数相同。*///标程：#include <iostream>#include <cstdio>#include <cmath>using namespace std;__int64  gcd(__int64 a,__int64 b){    return b==0?a:gcd(b,a%b);}int main(){//    freopen("a.txt","r",stdin);    int t,  i;    __int64 n, m, a, b;    cin >> t;    while(t --)    {        scanf("%I64d%I64d",&n,&m);             scanf("%I64d%I64d",&a,&b);        __int64  tmpx = a;        for(i = 1; i <= m; ++ i) tmpx *= 10;        tmpx = tmpx + b - a;        __int64 tmpy = 0;        for(i = 0; i < m; ++ i) tmpy = tmpy*10+9;        for(i = 1;i <= n; ++ i)  tmpy = tmpy*10;        __int64 x = gcd(tmpx,tmpy);        printf("%I64d %I64d\n",tmpx/x,tmpy/x);    }    return 0;}