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

3 1
5 0 2
5 1 6
4 2
1 2 7 9
5 6 7 9
0 0

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;
}