Best Coder #86 1001 Price List(大水题)

Price List

Accepts: 880
Submissions: 2184
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 262144/131072 K (Java/Others)
Problem Description

There are nnn shops numbered with successive integers from 111 to nnn in Byteland. Every shop sells only one kind of goods, and the price of the iii-th shop's goods is viv_ivi​​.

Every day, Byteasar will purchase some goods. He will buy at most one piece of goods from each shop. Of course, he can also choose to buy nothing. Back home, Byteasar will calculate the total amount of money he has costed that day and write it down on his account book.

However, due to Byteasar's poor math, he may calculate a wrong number. Byteasar would not mind if he wrote down a smaller number, because it seems that he hadn't used too much money.

Please write a program to help Byteasar judge whether each number is sure to be strictly larger than the actual value.

Input

The first line of the input contains an integer TTT (1≤T≤10)(1\leq T\leq 10)(1T10), denoting the number of test cases.

In each test case, the first line of the input contains two integers n,mn,mn,m (1≤n,m≤100000)(1\leq n,m\leq 100000)(1n,m100000), denoting the number of shops and the number of records on Byteasar's account book.

The second line of the input contains nnn integers v1,v2,...,vnv_1, v_2, ..., v_nv1​​,v2​​,...,vn​​ (1≤vi≤100000)(1\leq v_i\leq 100000)(1vi​​100000), denoting the price of the iii-th shop's goods.

Each of the next mmm lines contains an integer qqq (0≤q≤1018)(0\leq q\leq 10^{18})(0q1018​​), denoting each number on Byteasar's account book.

Output

For each test case, print a line with mmm characters. If the iii-th number is sure to be strictly larger than the actual value, then the iii-th character should be '1'. Otherwise, it should be '0'.

Sample Input
1
3 3
2 5 4
1
7
10000
Sample Output
001
#include <iostream>
#include <stdio.h>
#include <string.h>
#define N (1<<12)+5
#define M 12
using namespace std;
int main()
{
    //freopen("in.txt","r",stdin);
    int t;
    long long a,b,n,m,s;
    scanf("%d",&t);
    while(t--)
    {
        s=0;
        scanf("%lld%lld",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%lld",&a);
            s+=a;
        }
        for(int i=0;i<m;i++)
        {
            scanf("%lld",&b);
            if(b>s)
                printf("1");
            else
                printf("0");
        }
        printf("\n");
    }
    return 0;
}

 

posted @ 2016-08-06 21:31  勿忘初心0924  阅读(170)  评论(0编辑  收藏  举报