Write a program that will accept a fraction of the form N/D, where N is the numerator and D is the denominator and print the decimal representation. If the decimal representation has a repeating sequence of digits, indicate the sequence by enclosing it in brackets. For example, 1/3 = .33333333...is denoted as 0.(3), and 41/333 = 0.123123123...is denoted as 0.(123). Use xxx.0 to denote an integer. Typical conversions are:
1/3 = 0.(3) 22/5 = 4.4 1/7 = 0.(142857) 2/2 = 1.0 3/8 = 0.375 45/56 = 0.803(571428)
A single line with two space separated integers, N and D, 1 <= N,D <= 100000.
45 56
The decimal expansion, as detailed above. If the expansion exceeds 76 characters in length, print it on multiple lines with 76 characters per line.
0.803(571428)
就是这样一道比较简单的题目，是usaco上第二章的最后一道题目。（因为最近老师布置了usaco上的题目......如果看英文有问题，可以去nocow上看）
简单来说，这道题目就是一道类似高精小数的题目，只需要模拟除法就可以了。
但是为什么要推荐呢？因为特判和特殊情况很多......还有就是76个字符提一行。
可以试试自己做题考虑是否周到，能一次AC是很不容易的......
下面贴代码（我的代码很丑，自己参考）
#include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<map> using namespace std; int n,m,zhen; map <int,int> f; struct like { int w[100005]; }ans; int wa[10005],wo; void chu(int x,int y) { int i,j; zhen=x/y; x=x%y; f[x]=1; int t=1,anss=0; while(x!=0) { t++; x=x*10; ans.w[0]++; ans.w[ans.w[0]]=x/y; x=x%y; if(f[x]!=0) { anss=f[x]; break; } f[x]=t; } int p=0; while(zhen>0) { wo++; wa[wo]=zhen%10; zhen/=10; } for(i=wo;i>=1;--i) { printf("%d",wa[i]); p++; if(p==76) p=0,printf("\n"); } if(p==0) printf("0"),p++; printf(".");p++; if(p==76) p=0,printf("\n"); if(t==1) { printf("0"); printf("\n"); return ; } if(anss==0) { for(i=1;i<=ans.w[0];++i) { printf("%d",ans.w[i]); p++; if(p==76) p=0,printf("\n"); } if(p!=0) printf("\n"); return ; } for(i=1;i<=anss-1;++i) { printf("%d",ans.w[i]); p++; if(p==76) p=0,printf("\n"); } printf("(");p++; if(p==76) p=0,printf("\n"); for(i=anss;i<=t-1;++i) { printf("%d",ans.w[i]); p++; if(p==76) p=0,printf("\n"); } printf(")");p++; printf("\n"); } int main() { freopen("fracdec.in","r",stdin); freopen("fracdec.out","w",stdout); int i,j; scanf("%d%d",&n,&m); for(i=1;i<=10005;++i) ans.w[i]=-1; chu(n,m); return 0; }
usaco题目分享——Fractions to Decimals
原文：http://www.cnblogs.com/hhhhhhhhhh/p/5154721.html