2018 Multi-University Training Contest 4.Problem D. Nothing is Impossible（水题）

最新推荐文章于 2019-07-23 19:41:07 发布

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最新推荐文章于 2019-07-23 19:41:07 发布

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本文链接：https://blog.youkuaiyun.com/XxxxxM1/article/details/81353078

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本文探讨了一个关于学生如何选择解答试题以最大化胜算的问题。在已知每道题正确与错误选项数量的情况下，通过最优策略确定学生群体可以正确解答的最大题目数量。

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m students, including Kazari, will take an exam tomorrow.
The paper consists of exactly n problems, the i-th problem contains ai correct answers and bi incorrect answers, i.e. the i-th problem contains ai+bi candidates in total.
Each student should choose exactly one candidate as answer for each problem. If the answer to a certain problem is correct, then the student will get one point. The student who gets the most points wins.
Students only know the structure of the paper, but they are able to talk with each other during the exam. They decide to choose a subset S of all n problems, and they will only be able to submit answers on these problems.
They want to know the maximum size of S that the winner among them will solve all the problems in S if they take the optimal strategy.

For sample 1, students can choose S={1}，and we need at least 4 students to guarantee the winner solve the only problem.

For sample 2, students can choose S={1,2,3}, and we need at least 24 students to guarantee the winner solve these three problems, but if |S|=4, we need at least 96 students, which is more than 50.

Input

The first line of the input contains an integer T (1≤T≤100) denoting the number of test cases.
Each test case starts with two integers n,m (1≤n≤100,1≤m≤109), denoting the number of problems and the number of students. Each of next n lines contains two integers ai,bi （1≤bi≤100,ai=1）, indicating the number of correct answers and the number of incorrect answers of the i-th problem.

Output

For each test case, print an integer denoting the maximum size of S.

Sample Input

2 3 5 1 3 1 3 1 3 5 50 1 1 1 3 1 2 1 3 1 5

Sample Output

1 3

解题思路：按照最优策略，将正错比最高按照从高到低进行排序，依次求出排序后保持正确的人数，直到在下一次分配中无法满足剩余的正确人数完全覆盖正确和错误结果，则计数完毕。

#include<bits/stdc++.h>
using namespace std;
struct pp
{
    double z,c;
}p[103];
bool cmp(pp x,pp y)
{
    return x.z/x.c>y.z/y.c;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        double m;
        scanf("%d%lf",&n,&m);
        for(int i=0;i<n;i++)
           scanf("%lf%lf",&p[i].z,&p[i].c);
        sort(p,p+n,cmp);
        
        int sum=0;
        for(int i=0;i<n;i++)
        {
           if(m>=p[i].z+p[i].c)
           {
               m=(int)(m*p[i].z/(p[i].z+p[i].c)+0.1);
               sum++;
           }
           else
               break;
        }
        printf("%d\n",sum);
    }
    return 0;
}