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Fraction |
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| Accepted : 168 | Submit : 1061 | |
| Time Limit : 1000 MS | Memory Limit : 65536 KB | |
FractionProblem Description:Everyone has silly periods, especially for RenShengGe. It‘s a sunny day, no one knows what happened to RenShengGe, RenShengGe says that he wants to change all decimal fractions between 0 and 1 to fraction. In addtion, he says decimal fractions are too complicate, and set that [Math Processing Error] is much more convient than 0.33333... as an example to support his theory. So, RenShengGe lists a lot of numbers in textbooks and starts his great work. To his dissapoint, he soon realizes that the denominator of the fraction may be very big which kills the simplicity that support of his theory. But RenShengGe is famous for his persistence, so he decided to sacrifice some accuracy of fractions. Ok, In his new solution, he confines the denominator in [1,1000] and figure out the least absolute different fractions with the decimal fraction under his restriction. If several fractions satifies the restriction, he chooses the smallest one with simplest formation. InputThe first line contains a number T(no more than 10000) which represents the number of test cases. And there followed T lines, each line contains a finite decimal fraction x that satisfies [Math Processing Error] . OutputFor each test case, transform x in RenShengGe‘s rule. Sample Input3 Sample Output1/1 tipYou can use double to save x; |
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 using namespace std; 6 int gcd(int a,int b) 7 { 8 int c,t; 9 if(a<b) 10 { 11 t=a,a=b,b=t; 12 } 13 while(b) 14 { 15 c=a%b; 16 a=b; 17 b=c; 18 } 19 return a; 20 } 21 int main() 22 { 23 int i,j,T; 24 double s,minn; 25 scanf("%d",&T); 26 while(T--) 27 { 28 scanf("%lf",&s); 29 int a=0,b=1; 30 minn=s; 31 for(i=1; i<=1000; i++) //枚举1-1000的分母 32 { 33 j=s*i+0.5; //求出分子 34 double f=j*1.0/i; //计算此时分数的结果 35 double p=fabs(f-s); //与原来的数进行比较 36 if(minn>p) 37 { 38 minn=p; 39 a=j; 40 b=i; 41 } 42 } 43 int r=gcd(a,b); //求最大公约数，化简 44 printf("%d/%d\n",a/r,b/r); 45 } 46 return 0; 47 }
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原文地址：http://www.cnblogs.com/pshw/p/5572749.html
