Sample Input
76 25
5 43
1 397
Sample Output
76/25 = 3.04(0) 1 = number of digits in repeating cycle 5/43 = 0.(116279069767441860465) 21 = number of digits in repeating cycle 1/397 = 0.(00251889168765743073047858942065491183879093198992...) 99 = number of digits in repeating cycle输入分子和分母，找出来循环的小数部分
#include<cstdio>
#include<cstring>
#include<stack>
#include<iostream>
using namespace std;
#define M 3000
int main()
{
int a,b;
a = b = 0;
while(cin >> a >> b){//可以直接这样输入
printf("%d/%d = %d.",a,b,a/b);
int yushu[M] = {0},shang[M] = {0};
/// memset(shang, 0, 30);
//取余数，求初值后，循环的来计算
a %= b;
a *= 10;
for(int t = 0; t < M; t++) {
shang[t] = a/b;
a = a%b;
yushu[t] = a;
a *= 10;
}
//判断从哪一位开始循环。很巧妙
int i ,j;
for(i = 0; i < M; i++){
for(j = i + 1; j < M; j++) {
if(((shang[j] == shang[i]) && (yushu[j] == yushu[i]) )|| (i >= 100)){
goto a1;
}
}
}
a1:for(int m = 0; m < j; m++){
if(m == i) printf("(");
if(m<50) printf("%d",shang[m]);
else {printf("..."); break;}
}
printf(")\n");
printf(" %d = number of digits in repeating cycle\n\n",j-i);
}
return 0;
}
1 自己看了别人的代码，才学会了如何处理一个数的小数部分。模拟平时手算的步骤。
2 跳出两个循环可以用goto 。不要用break
3 发现提前定义一个变量M。这样就可以快速切换数组大小，方便debug
4 赋值直接用 int yushu[M] = {0},shang[M] = {0} 。如果用memset 有时候会出错
