POJ 1742 hdu 2844 Coins

题目链接：http://poj.org/problem?id=1742

http://acm.hdu.edu.cn/showproblem.php?pid=2844

题目分类：动态规划

代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>

using namespace std;
int n,m;
bool flag[];

struct P
{
    int a,b;

}point[];

int cmp(P X,P Y)
{
    return X.a<Y.a;
}

int main()
{
    while(scanf("%d %d",&n,&m)&&(n+m))
    {
        memset(point,,sizeof(point));
        memset(flag,,sizeof(flag));

        for(int i=;i<=n;i++)
        {
            scanf("%d",&point[i].a);
        }
        for(int i=;i<=n;i++)
        {
            scanf("%d",&point[i].b);
        }
        sort(point+,point+n+,cmp);
        int sum[];

        flag[]=;
        for(int i=;i<=n;i++)
        {
            memset(sum,,sizeof(sum));
            for(int k=point[i].a;k<=m;k++)
            {
                if(flag[k-point[i].a]&&sum[k-point[i].a]<point[i].b&&!flag[k])
                {
                    sum[k] = sum[k-point[i].a] + ;
                    flag[k] = ;
                }
            }
        }

        int ans=;
        for(int i=;i<=m;i++)
        {
            if(flag[i])
            {
                //printf("sss==   %d\n",i);
                ans++;
            }
        }
        printf("%d\n",ans);
    }
    return ;
}

#include<bits/stdc++.h>

using namespace std;

bool dp[];
int TmpVal[];
int val[];

int main()
{
    cin.sync_with_stdio(false);
    cout.sync_with_stdio(false);
    int n, m;
    while(cin>>n>>m)
    {
        if(n== && m==)
            break;
        memset(dp, , sizeof(dp));
        for(int i=;i<=n;i++)
            cin>>TmpVal[i];
        int N = ;
        for(int i=;i<=n;i++)
        {
            int num;
 cin>>num;
            int j = 1;
            while(j<=num)
            {
                val[++N] = TmpVal[i] * j;
                num -= j;
                j = (j<<1);
            }
            if(num)
            {
                val[++N] = TmpVal[i] * num;
            }
        }
 dp[0] = 1;
for(int i=;i<=N;i++)
        {
            for(int j=m;j>=val[i];j--)
            {
                if(dp[j-val[i]])
                    dp[j] = ;
            }
        }
        int ans = ;
        for(int i=;i<=m;i++)
            if(dp[i])
                ans++;
        printf("%d\n", ans);
    }
    return ;
}