poj2976(二分，最大值平均化模板）Dropping tests

/*浮点数的精度问题是通过多次循环实现的 (n-k)个数，最大化平均值*/ #include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int mn=1005;int n,m,a[mn],b[mn];double y[mn];bool iswork(double x){for(int i=0;i<n;++i)y[i]=a[i]-x*b[i];sort(y,y+n,greater<int>());double sum=0;for(int i=0;i<m;++i)sum+=y[i];return sum>=0.0;}void solve(){double l=0,r=1e9;int cnt=100;//100次的循环精度可以达到10^(-30)的精度范围while(cnt--){double mid=(l+r)/2;if(iswork(mid)) l=mid;else r=mid;}printf("%.0f\n",l*100);//在做题时printf最好用%f，否则可能出现一些莫名其妙的错误}int main(){while(~scanf("%d%d",&n,&m)&&(n||m)){m=n-m;for(int i=0;i<n;++i) scanf("%d",a+i);for(int i=0;i<n;++i) scanf("%d",b+i);solve();}return 0;}