poj 1742 Coins

類型：DP

題目：http://poj.org/problem?id=1742

來源：LouTiancheng@POJ

思路：用cnt數組記錄每個數據使用當前coin的次數，在次數限制範圍內不斷更新未達到的值 O(nm)

// poj 1742 Coins
// ac 768K 1532MS
#include <iostream>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define FORD(i,a,b) for(i = (a); i > (b); --i)
#define FORDE(i,a,b) for(i = (a); i >= (b); --i)
#define CLR(a,b) memset(a,b,sizeof(a))
#define PB(x) push_back(x)

const int MAXN = 122000;
const int MAXM = 0;
const int hash_size = 25000002;
const int INF = 0x7f7f7f7f;

int n, m;
int cnt[MAXN], val[MAXN], c[MAXN];
bool dp[MAXN];

int main() {
    int i, j, n;
    while(scanf("%d %d", &n, &m) == 2, n || m) {
        FORE(i, 1, n)
            scanf("%d", &val[i]);
        FORE(i, 1, n)
            scanf("%d", &c[i]);
        CLR(dp, false);
        dp[0] = true;
        int ans = 0;
        FORE(i, 1, n) {
            CLR(cnt, 0);
            FORE(j, val[i], m)
                if(dp[j - val[i]] && !dp[j] && cnt[j - val[i]] < c[i]) {
                    cnt[j] = cnt[j - val[i]] + 1;
                    ans++;
                    dp[j] = true;
                }
        }
        printf("%d\n", ans);
    }
    return 0;
}