【POJ】2976 Dropping tests（二分）

【题目链接】http://poj.org/problem?id=2976

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题目内容

    In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to beGiven 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 ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 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 15 0 25 1 64 21 2 7 95 6 7 90 0

Sample Output

83100

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).

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题目大意

给你两个数组a[],b[]，分别代表每次测试得到的分和总分，可以除掉其中的k对，要使sum(a[i])/sum(b[i])*100，也就是剩下的占分比例最大。

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解题思路

一开始想用贪心，但是找不出合适的规律尤其是数据中包括0，所以放弃了233.
直接二分删除后的sum(a)/sum(b),减去对这个答案贡献最小的几对，看能不能取到这个答案，在我们要找的最终答案的左侧都是可以取到的，右侧都不能。注意最后输出时四舍五入。

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AC代码

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<string>#include<algorithm>#include<math.h>#include<limits.h>#include<stack>#include<queue>#define LL long longusing namespace std;struct node{    LL a;    LL b;    double c;//cJ就代表这组数据对答案的贡献} p[1005];bool cmp(node x,node y){    return x.c<y.c;}int main(){    int n,k;    while(scanf("%d%d",&n,&k)!=EOF&&!(n==0&&k==0))    {        for(int i=0;i<n;i++)        {           scanf("%lld",&p[i].a);        }        for(int i=0;i<n;i++)        {           scanf("%lld",&p[i].b);        }        double left=0.0,right=1.1,mid,ans,x,y;        while(fabs(right-left)>1e-8)        {            mid=(right+left)/2;            for(int i=0;i<n;i++)            {               p[i].c=p[i].a-(double)p[i].b*mid;//a中超出b*mid的那部分就是这组数据对最终结果能造成的影响，正数是提高，负数是减少。            }            sort(p,p+n,cmp);            ans=0;            for(int i=k;i<n;i++)            {                ans+=p[i].c;            }            if(ans>=0)                left=mid;            else right=mid;        }        printf("%.0f\n",100*left);//浮点数输出会自动四舍五入    }    return 0;}