zoukankan      html  css  js  c++  java
  • Jam's balance set 暴力

    Jim has a balance and N weights. (1N20)(1≤N≤20) 
    The balance can only tell whether things on different side are the same weight. 
    Weights can be put on left side or right side arbitrarily. 
    Please tell whether the balance can measure an object of weight M.

    InputThe first line is a integer T(1T5)T(1≤T≤5), means T test cases. 
    For each test case : 
    The first line is NN, means the number of weights. 
    The second line are NN number, i'th number wi(1wi100)wi(1≤wi≤100) means the i'th weight's weight is wiwi. 
    The third line is a number MM. MM is the weight of the object being measured.
    OutputYou should output the "YES"or"NO".Sample Input

    1
    2
    1 4
    3
    2
    4
    5

    Sample Output

    NO
    YES
    YES
    
    
            
     

    Hint

    For the Case 1:Put the 4 weight alone
    For the Case 2:Put the 4 weight and 1 weight on both side
    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    #include<sstream>
    #include<algorithm>
    #include<queue>
    #include<vector>
    #include<cmath>
    #include<map>
    #include<stack>
    #include<set>
    #include<memory>
    #include<bitset>
    #include<string>
    #include<functional>
    using namespace std;
    typedef long long LL;
    typedef unsigned long long ULL;
    
    #define MAXN 23
    #define INF 0x3f3f3f3f
    /*
    给你一个数字M
    +-x +-y 能否==M
    */
    int a[MAXN], n, m;
    set<int> s;
    int main()
    {
        int T, tmp;
        scanf("%d", &T);
        while (T--)
        {
            s.clear();
            scanf("%d", &n);
            for (int i = 1; i <= n; i++)
                scanf("%d", &a[i]);
            //int t1, t2;
            for (int i = 1; i <= n; i++)
            {
                set<int>::iterator e = s.end();
                vector<int> t;
                for (set<int>::iterator it = s.begin(); it != s.end(); it++)
                {
                    if (!s.count(*it + a[i]))
                        t.push_back(*it + a[i]);
                    if (!s.count(abs(*it - a[i])))
                        t.push_back(abs(*it - a[i]));
                }
                s.insert(a[i]);
                for (int i = 0; i < t.size(); i++)
                    s.insert(t[i]);
            }
            scanf("%d", &m);
            while (m--)
            {
                scanf("%d", &tmp);
                if (s.count(tmp))
                    printf("YES
    ");
                else
                    printf("NO
    ");
            }
        }
    }
  • 相关阅读:
    Chapter 14 高级I/O
    UNP总结 Chapter 3 套接字编程简介
    UNP总结 Chapter 1 简介
    Chapter 13 守护进程
    Chapter 17 高级进程间通信
    Chapter 15 进程间通信
    实例详解JSP内置对象
    一个记录程序运行时间表的控件
    .Grove—— .Net下的ORM框架
    .Net 下的Wondows窗体常用项目
  • 原文地址:https://www.cnblogs.com/joeylee97/p/7395475.html
Copyright © 2011-2022 走看看