http://poj.org/problem?id=2976

通过二分 不断的移动左右边界 知道满足精度

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#define LL long long

using namespace std;

const double eps=1e-4;
const int N=1005;
double a[N];
double b[N];
double c[N];
bool cmp(double x,double y)
{
    return x>y;
}
int main()
{
    int n,k;
    while(scanf("%d %d",&n,&k)!=EOF)
    {
        if(n==0&&k==0)
        break;
        for(int i=0;i<n;++i)
        scanf("%lf",&a[i]);
        for(int i=0;i<n;++i)
        scanf("%lf",&b[i]);
        double l=0.0;
        double r=100.0;
        double mid;
        while(r-l>eps)
        {
            mid=(l+r)/2;
            for(int i=0;i<n;++i)
            c[i]=a[i]*100.0-b[i]*mid;
            sort(c,c+n,cmp);
            double sum=0;
            for(int i=0;i<n-k;++i)
            sum+=c[i];
            if(sum>0.0)
            l=mid;
            else
            r=mid;
        }
        printf("%d\n",int(l+0.5));
    }
    return 0;
}

  

posted on 2012-07-28 15:37  夜->  阅读(159)  评论(0编辑  收藏  举报