HDU5804--Price List
发布日期:2021-05-09 04:21:10 浏览次数:20 分类:博客文章

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Price List

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/131072 K (Java/Others)
Total Submission(s): 731    Accepted Submission(s): 417


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
 
(1T10)
, denoting the number of test cases.
In each test case, the first line of the input contains two integers 
n,m
 
(1n,m100000)
, 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
 
(1vi100000)
, denoting the price of the 
i
-th shop's goods.
Each of the next 
m
 lines contains an integer 
q
 
(0q1018)
, 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
13 32 5 41710000
 

Sample Output
001

解题思路:超过商品总价值就是用超了

源代码:

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;typedef long long ll;ll ans[100001];int main(){ int t; int n,m; int i; ll sum; ll temp; char s[100001]; scanf("%d",&t); while(t--) { memset(ans,0,sizeof(ans)); memset(s,0,sizeof(s)); sum = 0; scanf("%d%d",&n,&m); for(i = 0; i < n; i++) { scanf("%lld",&ans[i]); sum+=ans[i]; } for(i = 0; i < m; i++) { scanf("%lld",&temp); if(temp <= sum) s[i]='0'; else s[i]='1'; } printf("%s\n",s); } return 0;}

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路过按个爪印,很不错,赞一个!
[***.219.124.196]2025年05月07日 18时24分32秒