HDU 5616 Jam's balance（01背包）

题意:

给定1个天平,20个砝码wi≤100,可以放在左边,放在右边或者不放,问是否能称出k重量的物品

分析:

看起来很像搜索,其实是01背包
放在右边,就相当于称出了负数,加一个OFFSET=20∗100=2000就好了

代码:

//
// Created by TaoSama on 2016-01-30
// Copyright (c) 2015 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, w[45], dp[4005];
const int OFF = 2000;

int main() {
#ifdef LOCAL
    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
    ios_base::sync_with_stdio(0);

    int t; scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) scanf("%d", w + i);
        memset(dp, false, sizeof dp);
        dp[OFF] = true;
        for(int i = 1; i <= n; ++i) {
            for(int j = OFF << 1; j >= w[i]; --j)
                dp[j] |= dp[j - w[i]];
            for(int j = 0; j <= (OFF << 1) - w[i]; ++j)
                dp[j] |= dp[j + w[i]];
        }
        int q; scanf("%d", &q);
        while(q--) {
            long long x; scanf("%I64d", &x);
            if(x < 0 || x > 2000 || !dp[x + OFF]) puts("NO");
            else puts("YES");
        }
    }
    return 0;
}