I NEED A OFFER!

本文介绍了一个留学申请场景下的数学问题,旨在帮助申请者在有限的资金内最大化获得至少一份offer的概率。通过动态规划算法求解最优选择方案,适用于计算机科学领域的算法学习与实践。

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K - I NEED A OFFER!

Status
Description
Speakless很早就想出国,现在他已经考完了所有需要的考试,准备了所有要准备的材料,于是,便需要去申请学校了。要申请国外的任何大学,你都要交纳一定的申请费用,这可是很惊人的。Speakless没有多少钱,总共只攒了n万美元。他将在m个学校中选择若干的(当然要在他的经济承受范围内)。每个学校都有不同的申请费用a(万美元),并且Speakless估计了他得到这个学校offer的可能性b。不同学校之间是否得到offer不会互相影响。“I NEED A OFFER”,他大叫一声。帮帮这个可怜的人吧,帮助他计算一下,他可以收到至少一份offer的最大概率。(如果Speakless选择了多个学校,得到任意一个学校的offer都可以)。 
 
Input
输入有若干组数据,每组数据的第一行有两个正整数n,m(0<=n<=10000,0<=m<=10000) 
后面的m行,每行都有两个数据ai(整型),bi(实型)分别表示第i个学校的申请费用和可能拿到offer的概率。 
输入的最后有两个0。 
 
Output
每组数据都对应一个输出,表示Speakless可能得到至少一份offer的最大概率。用百分数表示,精确到小数点后一位。 
 
Sample Input
10 3
4 0.1
4 0.2
5 0.3
0 0 
 
Sample Output
44.0%


           
Hint
 You should use printf("%%") to print a '%'. 


AC代码:

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
int main()
{
	double dp[10001],t;
	int n,m,i,j,k;
	while(scanf("%d%d",&n,&m)&&(m||n))
	{
		for(i=0;i<10001;++i)
		{
			dp[i]=1;
		}
		for(i=0;i<m;++i)
		{
			scanf("%d%lf",&k,&t);
			for(j=n;j>=k;--j)
			{
				dp[j]=min(dp[j],dp[j-k]*(1-t));
			}
		}
		printf("%0.1lf%%\n",(1-dp[n])*100);
	}
	return 0;
}

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最新发布
05-12
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