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)
PROGRAM NAME: fracdec
INPUT FORMAT
A single line with two space separated integers, N and D, 1 <= N,D <= 100000.
SAMPLE INPUT (file fracdec.in)
45 56
OUTPUT FORMAT
The decimal expansion, as detailed above. If the expansion exceeds 76 characters in length, print it on multiple lines with 76 characters per line.
SAMPLE OUTPUT (file fracdec.out)
0.803(571428)
题解：就是模拟除法嘛。我们用一个数组记录余数，如果一个已经出现过的余数再次出现，那么循环节就出现了。循环节是从余数上次出现的位置到余数此次出现的位置的前一个位置为止。如果余数为零，那么输出小数部分。如果输入的第一个数能整除第二个数，则不会产生小数部分，那么输出整数部分，然后输出小数点之后再输出一个0即可。
View Code
1 /* 2 ID:spcjv51 3 PROG:fracdec 4 LANG:C 5 */ 6 #include<stdio.h> 7 #include<stdlib.h> 8 #include<string.h> 9 long f[100005]; 10 int s[100005]; 11 int main(void) 12 { 13 freopen("fracdec.in","r",stdin); 14 freopen("fracdec.out","w",stdout); 15 long flag,n,d,ans,t,i,sum; 16 char ss[20]; 17 scanf("%ld%ld",&n,&d); 18 if(n%d==0) 19 printf("%ld.0\n",n/d); 20 else 21 { 22 printf("%ld.",n/d); 23 sprintf(ss,"%ld",n/d); 24 sum=strlen(ss)+1; 25 memset(f,0,sizeof(f)); 26 flag=1; 27 ans=0; 28 while(flag) 29 { 30 ans++; 31 t=n%d; 32 if(f[t]) 33 { 34 sum++; 35 for(i=1; i<ans; i++) 36 { 37 sum++; 38 if(i==f[t]) 39 printf("(%d",s[i]); 40 else 41 printf("%d",s[i]); 42 if(sum%76==0) printf("\n"); 43 } 44 printf(")\n"); 45 flag=0; 46 break; 47 } 48 else 49 { 50 f[t]=ans; 51 n=t*10; 52 s[ans]=n/d; 53 } 54 if(t==0) 55 { 56 for(i=1; i<ans; i++) 57 { 58 sum++; 59 printf("%d",s[i]); 60 if(sum%76==0) 61 printf("\n"); 62 } 63 printf("\n"); 64 flag=0; 65 break; 66 } 67 68 } 69 } 70 return 0; 71 }