poj 2976 Dropping tests 二分搜索+精度处理

Dropping tests

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8349		Accepted: 2919

Description

In a certain course, you take n tests. If you get a_i out of b_i questions correct on test i, your cumulative average is defined to be

Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.

Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .

Input

The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k <n. The second line contains n integers indicating a_i for all i. The third line contains n positive integers indicating b_i for all i. It is guaranteed that 0 ≤a_i ≤ b_i ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.

Output

For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.

Sample Input

Sample Output

83
100

Hint

To avoid ambiguities due to rounding errors, the judge tests have been constructed so that all answers are at least 0.001 away from a decision boundary (i.e., you can assume that the average is never 83.4997).

Source

Stanford Local 2005

分析：这道题目要是看了01规划，大水题一道，

01规划：http://www.cnblogs.com/perseawe/archive/2012/05/03/01fsgh.html

不过最后要注意的是因为要四舍五入，所以不能直接对整数二分，最后输出+0.005就是为了四舍五入

#include<cstdio>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int mod = 1000000007;
const double eps = 1e-10;
const int inf = 0x3f3f3f3f;
struct Node{
   int a,b;
   double temp;
}node[1005];
 int n,k;
bool cmp(Node a ,Node  b)
{
    return a.temp>b.temp;
}
int ok(double mid)
{
    for(int i=1;i<=n;i++)
        node[i].temp=node[i].a*1.0-mid*node[i].b;
    sort(node+1,node+n+1,cmp);
    double sum=0;
    for(int i=1;i<=n-k;i++)
          sum+=node[i].temp;
    return sum>=0;
}
int main()
{
    while(~scanf("%d %d",&n,&k))
    {
        if(n==0&&k==0) return 0;
        for(int i=1;i<=n;i++)
            scanf("%d",&node[i].a);
        for(int i=1;i<=n;i++)
            scanf("%d",&node[i].b);
        double l=0,r=1,mid;
        while(r-l>1e-5)
        {
            mid=(l+r)/2;
            if(ok(mid))
                l=mid;
            else
                r=mid;
        }
        printf("%d
",int(100*(l+0.005))); //四舍五入
    }
    return 0;
}