Title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1203
Title:
With $n in hand, there are m colleges and universities with a chance of admission. The application fee is a and the admission probability is b. Let's find the maximum probability of getting at least one offer.
Explanation:
To find the maximum probability of at least one offer, we can convert it to the minimum probability that we can't get one offer first.
dp[i] stores the probability that the smallest offer under I application fee can not be obtained. Here, the state transition equation is multiplication. In this case, because it is probability (real data), our dp [] must be initialized to 1, and ans must also be initialized to 1.
There is also a special case to note: n is 0, but there are schools with application fees of 0, this special treatment is OK.
Code:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <list>
#include <map>
#define P(x) x>0?x:0
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn=1e4+5;
const int maxm=1e4+5;

int n,m;
double dp[maxn];
double ans;
struct node
{
    int a;
    double b;
    node(int aa=0,double bb=0):a(aa),b(bb){}
    friend bool operator < (node n1,node n2)
    {
        return n1.a<n2.a;
    }
}item[maxm];

void init()
{
    for(int i=0;i<maxn;i++)
    {
        dp[i]=1;
    }
    ans=1;
}

int main()
{
    while(~scanf("%d%d",&n,&m)&&(n||m))
    {
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d%lf",&item[i].a,&item[i].b);
            item[i].b=1-item[i].b;//We let item[i].b save the probability of not being admitted
        }
        if(n==0)
        {
            for(int i=1;i<=m;i++)
            {
                if(item[i].a==0)
                    ans*=item[i].b;
            }
            printf("%.1f%%\n",(1-ans)*100);
            continue;
        }
        for(int i=1;i<=m;i++)
        {
            for(int j=n;j>=item[i].a;j--)
            {
                dp[j]=min(dp[j-item[i].a]*item[i].b,dp[j]);
            }
        }
        for(int i=1;i<=n;i++)
            ans=min(ans,dp[i]);
        printf("%.1f%%\n",(1-ans)*100);
    }
    return 0;
}