POJ 1742 dp

题意

传送门 POJ 1742

多重部分和问题

$dp[i+1][j]:=用前i种数加和得到j时第i种数最多能剩余多少个（不能加和得到j的情况下为-1）$

实现上重复利用数组，最终满足 $dp[i]\geq 0,i>0$ 的 $i$ 的数量即答案。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <queue>
#define min(a,b)    (((a) < (b)) ? (a) : (b))
#define max(a,b)    (((a) > (b)) ? (a) : (b))
#define abs(x)    ((x) < 0 ? -(x) : (x))
#define INF 0x3f3f3f3f
#define delta 0.85
#define eps 1e-3
#define PI 3.14159265358979323846
#define MAX_N 105
#define MAX_M 100005
using namespace std;
typedef pair<int, int> P;
int N, M;
int A[MAX_N], C[MAX_N];
int dp[MAX_M];

int main(){
	while(~scanf("%d%d", &N, &M) && (N | M)){
		memset(dp, -1, sizeof(dp));
		for(int i = 0; i < N; i++) scanf("%d", A + i);
		for(int i = 0; i < N; i++) scanf("%d", C + i);
		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 res = 0;
		for(int i = 1; i <= M; i++) if(dp[i] >= 0) ++res;
		printf("%d\n", res);
	}
	return 0;
}