Price List

时间:2021-09-07 18:07:26

Price List

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 296    Accepted Submission(s): 192

Problem Description
There are nPrice List

shops numbered with successive integers from 1Price List

to nPrice List

in Byteland. Every shop sells only one kind of goods, and the price of the iPrice List

-th shop's goods is vPrice ListiPrice ListPrice List

.

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 TPrice List

(1≤T≤10)Price List

, denoting the number of test cases.

In each test case, the first line of the input contains two integers n,mPrice List

(1≤n,m≤100000)Price List

, denoting the number of shops and the number of records on Byteasar's account book.

The second line of the input contains nPrice List

integers vPrice List1Price List,vPrice List2Price List,...,vPrice ListnPrice ListPrice List

(1≤vPrice ListiPrice List≤100000)Price List

, denoting the price of the iPrice List

-th shop's goods.

Each of the next mPrice List

lines contains an integer qPrice List

(0≤q≤10Price List18Price List)Price List

, denoting each number on Byteasar's account book.

Output
For each test case, print a line with mPrice List

characters. If the iPrice List

-th number is sure to be strictly larger than the actual value, then the iPrice List

-th character should be '1'. Otherwise, it should be '0'.

Sample Input
1
3 3
2 5 4
1
7
10000
Sample Output
001
Source
Recommend
wange2014   |   We have carefully selected several similar problems for you:  5808 5807 5806 5805 5804 
唉,昨晚只做出来一个题。
代码:
#include <iostream>
#include <cstdio> using namespace std; int main()
{
int t;
int n,m;
int v=;;
int sum=;
int jilu[];
int jieguo[];
scanf("%d",&t);
while(t--){
sum=;
scanf("%d %d",&n,&m);
for(int i=;i<n;i++){
scanf("%d",&v);
sum+=v;
}
for(int j=;j<m;j++){
scanf("%d",&jilu[j]);
}
for(int j=;j<m;j++){
if(jilu[j]>sum){
printf("");
}else{
printf("");
}
}
printf("\n"); }
return ;
}