暑假选拔赛01 ---- Chufa

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Problem Description

给出两个正整数m、n，输出m/n的精确值。
如果m/n是整数或有限小数，那么直接输出它的值。
如果m/n是无限小数，那么输出小数点后到第一个最小循环节为止，并用”()”把最小循环节括起来。
比如，4/2=2，输出”2”；1/5=0.2，输出”0.2”；5/6=0.833333…，输出”0.8(3)”。

Input

输入第一行为一个正整数T，表示有T组测试数据。
接下去T行，每行为一组数据。每行两个个正整数m、n，含义如上。
1<=m<=6,1<=n<=8。

Output

每个样例一行，输出答案。

Sample Input

3
4 2
1 5
5 6

Sample Output

2
0.2
0.8(3)

解题思路

总共n*m=48种情况,直接打表!

参考代码

#include <stdio.h>const char* p[8][6] = {"1","2","3","4","5","6","0.5","1","1.5","2","2.5","3","0.(3)","0.(6)","1","1.(3)","1.(6)","2","0.25","0.5","0.75","1","1.25","1.5","0.2","0.4","0.6","0.8","1","1.2","0.1(6)","0.(3)","0.5","0.(6)","0.8(3)","1","0.(142857)","0.(285714)","0.(428571)","0.(571428)","0.(714285)","0.(857142)","0.125","0.25","0.375","0.5","0.625","0.75"};int main(){    int T;    scanf("%d",&T);    while (T--){        int n,m;        scanf("%d %d",&m,&n);        printf("%s\n",p[n-1][m-1]);    }    return 0;}

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