题目链接：

Price List

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 262144/131072 K (Java/Others)

Problem Description

There are n shops numbered with successive integers from 1 to n in Byteland. Every shop sells only one kind of goods, and the price of the i-th shop's goods is vi.

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 T (1≤T≤10), denoting the number of test cases.

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

The second line of the input contains n integers v1,v2,...,vn (1≤vi≤100000), denoting the price of the i-th shop's goods.

Each of the next m lines contains an integer q (0≤q≤1018), denoting each number on Byteasar's account book.

Output

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

Sample Input

1
3 3
2 5 4
1
7
10000

Sample Output

001

题意：

问是否一定是记多了？

思路：

水题;

AC代码：

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┆　　┃　 AC代马 　　┣┓┆

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┆　　　┃┫┫　┃┫┫ ┆

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************************************************ */ 

#include <iostream>

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

//#include <bits/stdc++.h>

#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)

#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {

    char CH; bool F=false;

    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());

    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());

    F && (num=-num);

}

int stk[70], tp;

template<class T> inline void print(T p) {

    if(!p) { puts("0"); return; }

    while(p) stk[++ tp] = p%10, p/=10;

    while(tp) putchar(stk[tp--] + '0');

    putchar('\n');

}

const LL mod=1e9+7;

const double PI=acos(-1.0);

const int inf=1e9;

const int N=1e6+10;

const int maxn=2e3+14;

const double eps=1e-12;

int main()

{

        int t;

        read(t);

        while(t--)

        {

            int n,m,x,mmax=0;

            LL sum=0,temp;

            read(n);read(m);

            For(i,1,n)read(x),sum=sum+x;

            //sort(a+1,a+n+1);

            while(m--)

            {

                read(temp);

                if(temp>sum)printf("1");

                else printf("0");

            }

            printf("\n");

        }

        return 0;

}

巴特西

hdu-5804 Price List(水题)

Price List

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