HDU 1203 貪心或者DP

             貪心策略是申請offer需要的錢越少越好，而成功率越高越好，所以按照需要的錢和成功率的比值從小到大排序，依次選擇。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 10010
using namespace std;
struct node
{
    int money;
    double p,r;
    bool operator<(const node &th) const
    {
        return th.r>r;
    }
};
node  chan[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF&&(m+n))
    {
        for(int i=0; i<n; i++)
        {
            scanf("%d%lf",&chan[i].money,&chan[i].p);
            chan[i].r=chan[i].money/chan[i].p;
        }
        sort(chan,chan+n);
        int sum=m;
        double rate=1;
        for(int i=0; i<n; i++)
        {
            if(sum>=chan[i].money)
            {
                sum-=chan[i].money;
                rate=rate*(1-chan[i].p);
            }
        }
        printf("%.1lf%%\n",(1-rate)*100);
    }
    return 0;
}

         下面是按照0-1揹包思路做的，

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define MAXN 10010
using namespace std;
struct node
{
    int money;
    double p;
};
node  chan[MAXN];
int main()
{
    //freopen("in.txt","r",stdin);
    int m,n;
    while(scanf("%d%d",&m,&n)!=EOF&&(m+n))
    {
        for(int i=0; i<n; i++)
            scanf("%d%lf",&chan[i].money,&chan[i].p);
        double d[MAXN];
        for(int i=0;i<=m;i++)
            d[i]=1;
        for(int i=0; i<n; i++)
        {
            for(int j=m; j>=chan[i].money; j--)
            {
                if(d[j]>d[j-chan[i].money]*(1-chan[i].p))
                    d[j]=d[j-chan[i].money]*(1-chan[i].p);
            }
        }
     printf("%.1lf%%\n",(1-d[m])*100);
    }
    return 0;
}