多重部分和 poj1742

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.
You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

多重部分和问题：
有n中大小不同的数字，每种c[i]个，判断这些数字之中能否选出若干个使其和为K
此题是让求K<=m时，有多少个解

一个一般性的代码如下

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int n, m;
int a[102];
int c[102];
int dp[102][100002];

int main(int argc, char const *argv[])
{
    //freopen("input.txt","r",stdin);
    while(scanf("%d %d",&n,&m) != EOF && (n != 0 && m != 0)) {
        for(int i = 0; i < n; i++) {
            scanf("%d",&a[i]);
        }
        for(int i = 0; i < n; i++) {
            scanf("%d",&c[i]);
        }
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= m; j++) {
                for(int k = 0; k <= c[i] && k * a[i] <= j;k++) {
                    dp[i+1][j] = dp[i+1][j]| dp[i][j-k*a[i]];
                }
            }
        }
        int ans = 0;
        for(int i = 1; i <= m; i++) {
            if(dp[n][i] > 0) {
                ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}

若用dp[i+1][j]表示用前i个数相加和为j时第i种数最多能剩余几个（不能得到和为-1）

可得代码如下

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

int n, m;
int a[102];
int c[102];
int dp[100002];

int main(int argc, char const *argv[])
{
    //freopen("input.txt","r",stdin);
    while(scanf("%d %d",&n,&m) != EOF && (n != 0 && m != 0)) {
        for(int i = 0; i < n; i++) {
            scanf("%d",&a[i]);
        }
        for(int i = 0; i < n; i++) {
            scanf("%d",&c[i]);
        }
        memset(dp, -1, sizeof(dp));
        dp[0] = 0;
        for(int i = 0; i < n; i++) {
            for(int j = 0; j <= m; j++) {
                if(dp[j] >= 0) {
                    dp[j] = c[i];
                }
                else if(j < a[i] || dp[j - a[i]] <= 0) {
                    dp[j] = -1;
                }
                else {
                    dp[j] = dp[j - a[i]] - 1;
                }
            }
        }
        int ans = 0;
        for(int i = 1; i <= m; i++) {
            if(dp[i] >= 0) {
                ans++;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}