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:
 
/************************************************
┆ ┏┓   ┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃       ┃ ┆
┆┃   ━   ┃ ┆
┆┃ ┳┛ ┗┳ ┃ ┆
┆┃       ┃ ┆
┆┃   ┻   ┃ ┆
┆┗━┓   ┏━┛ ┆
┆  ┃   ┃  ┆      
┆  ┃   ┗━━━┓ ┆
┆  ┃  AC代马   ┣┓┆
┆  ┃    ┏┛┆
┆  ┗┓┓┏━┳┓┏┛ ┆
┆   ┃┫┫ ┃┫┫ ┆
┆   ┗┻┛ ┗┻┛ ┆
************************************************ */ #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;
}

  

hdu-5804 Price List(水题)的更多相关文章

  1. hdu 1106:排序(水题,字符串处理 + 排序)

    排序 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submissi ...

  2. HDU 4950 Monster (水题)

    Monster 题目链接: http://acm.hust.edu.cn/vjudge/contest/123554#problem/I Description Teacher Mai has a k ...

  3. HDU 4813 Hard Code 水题

    Hard Code Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.act ...

  4. HDU 4593 H - Robot 水题

    H - RobotTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.act ...

  5. HDOJ/HDU 2560 Buildings(嗯~水题)

    Problem Description We divide the HZNU Campus into N*M grids. As you can see from the picture below, ...

  6. HDOJ(HDU) 1859 最小长方形(水题、、)

    Problem Description 给定一系列2维平面点的坐标(x, y),其中x和y均为整数,要求用一个最小的长方形框将所有点框在内.长方形框的边分别平行于x和y坐标轴,点落在边上也算是被框在内 ...

  7. HDU - 1716 排列2 水题

    排列2 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submis ...

  8. HDU—2021-发工资咯(水题,有点贪心的思想)

    作为杭电的老师,最盼望的日子就是每月的8号了,因为这一天是发工资的日子,养家糊口就靠它了,呵呵  但是对于学校财务处的工作人员来说,这一天则是很忙碌的一天,财务处的小胡老师最近就在考虑一个问题:如果每 ...

  9. hdu 5753 Permutation Bo 水题

    Permutation Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 Description There are two sequen ...

随机推荐

  1. C# 调用API接口处理公共类 自带JSON实体互转类

    using System; using System.Collections.Generic; using System.IO; using System.Linq; using System.Net ...

  2. mysql中UNIX_TIMESTAMP()函数和php中time()函数的区别

    http://tech.ddvip.com/2009-01/1231392775105351.html mysql 中:UNIX_TIMESTAMP(), UNIX_TIMESTAMP(date) 若 ...

  3. Hibernate学习六----------CRUD

    © 版权声明:本文为博主原创文章,转载请注明出处 实例 1.项目结构 2.pom.xml <project xmlns="http://maven.apache.org/POM/4.0 ...

  4. CentOS下安装python3.x版本

    现在python都到了3.x版本,但是centos中自带的python仍然是2.7版本的,所以想把python换成3.x版本的. 但是这个地方有个坑,你要是直接编译安装了python3.x之后,估计你 ...

  5. 读懂JVM垃圾收集日志

    JVM垃圾收集搞懂原理很重要,但是连垃圾收集的日志都不懂,这tmd还分析个屁的问题啊,典型的空有理论知识,动手实践为零.本篇就来具体的学习一下JVM中的垃圾收集日志. 第一步,垃圾收集的选项 图1 J ...

  6. uboot之bootm以及go命令的实现

    本文档简单介绍了uboot中用于引导内核的命令bootm的实现,同时分析了uImage文件的格式,还简单看了一下uboot下go命令的实现 作者: 彭东林 邮箱: pengdonglin137@163 ...

  7. nginx - KeepAlive详细解释

    最近工作中遇到一个问题,想把它记录下来,场景是这样的: 从上图可以看出,用户通过Client访问的是LVS的VIP, VIP后端挂载的RealServer是Nginx服务器. Client可以是浏览器 ...

  8. [转]浅谈Flash Socket通信安全沙箱

    用过Flash socket的同学都知道,Flash socket通讯有安全沙箱问题.就是在Flash Player发起socket通信时,会向服务端获取安全策略,如果得不到服务端响应,flash将无 ...

  9. OpenCV 中的三大数据类型:CvMat 类型

    前言 本文将介绍 OpenCV 中的矩阵结构 CvMat 并提供几个很常用的矩阵使用方法. 更多的矩阵处理函数还请参阅相关资料. CvMat 的类型定义 typedef struct CvMat { ...

  10. android Bluetooth-蓝牙

    bluetooth 一.开启蓝牙 1.获取BluetoothAdapter BluetoothAdapter.getDefaultAdapter() 2.判断手机设备是否 有蓝牙模块 3.开启蓝牙设备 ...