poj 1079 Ratio

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//poj 1079 Ratio/*维护最小误差，枚举每个分母。注意两点：1.输入可能可以约分，这个有案例2.如果出现同分母两个分子都一样更精确，取分子大的*/#include <iostream>#include <algorithm>#include <cmath>using namespace std;const int inf = 1<<28;int n,m;#define eps 0.0000000001int gcd(int n,int m){    if (n<m) swap(n,m);    int r;    while (r=n%m)    {          n=m;          m=r;    }    return m;}int main(){    while (scanf("%d%d",&n,&m)!=EOF)    {          int g=gcd(n,m);          n/=g,m/=g;          double delta = inf*1.0,d;          bool flag;          for (int i=1;i<m;i++)          {              flag=false;              int b=n*i/m;              d=(double)n/m-(double)b/i;                  if( d<delta-eps)              {                   delta=d;                   flag=true;              }              ++b;              d=(double)b/i-(double)n/m;              if (d<delta-eps || flag && fabs(d-delta)<eps)              {                   delta=d;                   printf("%d/%d/n",b,i);              }              else if (flag) printf("%d/%d/n",b-1,i);                     }          printf("%d/%d/n/n",n,m);              }    system("pause");    return 0;}