PAT 1048 Find Coins[比较]
1048 Find Coins (25 分)
Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 105 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.
Input Specification:
Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (≤105, the total number of coins) and M (≤103, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print in one line the two face values V1 and V2 (separated by a space) such that V1+V2=Mand V1≤V2. If such a solution is not unique, output the one with the smallest V1. If there is no solution, output No Solution instead.
Sample Input 1:
8 15
1 2 8 7 2 4 11 15
Sample Output 1:
4 11
Sample Input 2:
7 14
1 8 7 2 4 11 15
Sample Output 2:
No Solution
题目大意:给出n个数,并给出一个两个数的和,判断是否存在这样的两个数,如果有相同的,输出V1较小的结果。
#include <iostream>
#include <algorithm>
#include <vector>
#include<string.h>
#include<string>
#include<cstdio>
using namespace std; int a[];
int main()
{
int n,total;
cin>>n>>total;
for(int i=;i<n;i++){
cin>>a[i];
}
int f=-,s=-;
sort(a,a+n);//从小到大排列。
bool flag=true;
for(int i=;i<n-&&flag;i++){
for(int j=i+;j<n;j++){
if(a[i]+a[j]==total){
f=a[i];
s=a[j];
flag=false;break;
}else if(a[i]+a[j]>total)break;
}
}
if(f!=-&&s!=-)
cout<<f<<" "<<s;
else
cout<<"No Solution";
return ;
}
//一开始写成这个样子,牛客网上通过60%,PAT上有两个测试点没通过,都是因为运行超时,不知如何解决。
下是柳神的解答,实在是叹为观止。

1.使用数组记录数字出现的个数,
2硬币面值不超过500!!!。。
PAT 1048 Find Coins[比较]的更多相关文章
- PAT 1048. Find Coins
two sum题目,算是贪婪吧 #include <cstdio> #include <cstdlib> #include <vector> #include &l ...
- PAT 解题报告 1048. Find Coins (25)
1048. Find Coins (25) Eva loves to collect coins from all over the universe, including some other pl ...
- PAT甲 1048. Find Coins (25) 2016-09-09 23:15 29人阅读 评论(0) 收藏
1048. Find Coins (25) 时间限制 50 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Eva loves t ...
- PAT 甲级 1048 Find Coins (25 分)(较简单,开个数组记录一下即可)
1048 Find Coins (25 分) Eva loves to collect coins from all over the universe, including some other ...
- 浙大pat 1048 题解
1048. Find Coins (25) 时间限制 50 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Eva loves t ...
- 1048 Find Coins (25 分)
1048 Find Coins (25 分) Eva loves to collect coins from all over the universe, including some other p ...
- PAT 甲级 1048 Find Coins
https://pintia.cn/problem-sets/994805342720868352/problems/994805432256675840 Eva loves to collect c ...
- PAT Advanced 1048 Find Coins (25 分)
Eva loves to collect coins from all over the universe, including some other planets like Mars. One d ...
- PAT Advanced 1048 Find Coins (25) [Hash散列]
题目 Eva loves to collect coins from all over the universe, including some other planets like Mars. On ...
随机推荐
- 10 部署应用程序和applet
跳过 09 Swing用户界面组件 JAR文件 在将应用程序进行打包时, 使用者一定希望仅提供给其一个单独的文件, 而不是一个含有大量类文件的目录, Java归档(JAR)文件就是为此目的而设计的. ...
- python ascii codec can't decode
提示错误: UnicodeDecodeError: 'ascii' codec can't decode byte 0xe5 in position 240: ordinal not in range ...
- 重新=》easyui DataGrid是否可以动态的改变列显示的顺序
$.extend($.fn.datagrid.methods,{ columnMoving: function(jq){ return jq.each(function(){ var target = ...
- Android实现时间轴
昨天群里有讨论时间轴的项目,没有接触过,以为非常吊,研究之后才知道表面都是忽悠人的,使用listview就能实现了,也没有什么新奇的东西 废话少说,直接上图 图片和文字都能够私人订制 没什么好说的,直 ...
- db2 improt from coldel0x7c
db2 load from "C:\20110816\20110816_BankEnterpri seCA.txt" OF del modified by coldel0x7c r ...
- 【代码备份】NLM插值
文件路径: main.m: %% 测试函数 clc,clear all,close all; %输入的原始小图 ima_ori=double(imread('F:\Users\****n\Docume ...
- 【BZOJ2792】[Poi2012]Well 二分+双指针法
[BZOJ2792][Poi2012]Well Description 给出n个正整数X1,X2,...Xn,可以进行不超过m次操作,每次操作选择一个非零的Xi,并将它减一. 最终要求存在某个k满足X ...
- Android - ViewPager实现Gallery效果
RelativeLayout viewPagerContainer = (RelativeLayout) headerView.findViewById(R.id.content_pager_layo ...
- LAMP集群项目三 配置业务服务器
安装MySQL 参考脚本:CentOS6.5一键安装MySQL5.5.32(源码编译) 在备份服务器上配置rsync推送任务 在备份服务器上配置 /etc/rsyncd.conf #在所有的客户端都 ...
- DOS和BAT批量提取修改文件名
DOS命令窗口:开始-cmd-回车,进入DOS命令窗口 案例一.获取文件名 dir 1.输入"文件所在盘",回车,如: d: 2.输入"cd 文件夹位置",回车 ...