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
- 哪一个不是EL定义的隐式对象?(选择1项)
哪一个不是EL定义的隐式对象?(选择1项) A cookie B.pageContext C.attributes D initParam 解答:C 1)pageContext:JSP 页的上下文.它 ...
- java 多线程 1 “常用的实现多线程的2种方式”:Thread 和 Runnable
转载系列自http://www.cnblogs.com/skywang12345/p/java_threads_category.html 当使用第一种方式(继承Thread的方式)来生成线程对象时, ...
- 【BZOJ】1101: [POI2007]Zap(莫比乌斯+分块)
http://www.lydsy.com/JudgeOnline/problem.php?id=1101 无限膜拜数论和分块orz 首先莫比乌斯函数的一些性质可以看<初等数论>或<具 ...
- hdu 2527:Safe Or Unsafe(数据结构,哈夫曼树,求WPL)
Safe Or Unsafe Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)To ...
- 【Python】用文本打印树
From:http://zhidao.baidu.com/link?url=O8U5TynGBMojDw2iFhlghPPf5_ZE1X8CAQMrK19pv-KxhvKCc6Z2yzsoQaukgN ...
- 项目期复习总结2:Table, DIV+CSS,标签嵌套规则
文件夹: 1.表格的意义,含义? 2.表格有哪些元素? 3.表格布局,表格布局的优缺点 4.行元素,块元素的差别? 5.标签的合理嵌套及标签的语义性 ① 表格的意义,含义? 表格应该用来展现那些适合表 ...
- OSG简单测试框架
#include <osgDB/ReadFile> #include <osgDB/FileUtils> #include <osg/ArgumentParser> ...
- MySQL- INSTR 函数的用法
测试数据库: MYSQL数据库 INSTR(STR,SUBSTR) 在一个字符串(STR)中搜索指定的字符(SUBSTR),返回发现指定的字符的位置(INDEX); STR 被搜索的字符串 SUBST ...
- php实现简单的权限管理
今天主要来实现一个权限管理系统,它主要是为了给不同的用户设定不同的权限,从而实现不同权限的用户登录之后使用的功能不一样,首先先看下数据库 总共有5张表,qx_user,qx_rules和qx_jues ...
- 深搜———ZOJ 1004:anagrams by stack
细节问题各种虐!! 其实就是简单的一个深搜 看成二叉树来理解:每个节点有两个枝:入栈和出栈. 剪枝操作:只有当栈顶元素和当前位置的目标字符相同时才出栈,否则就不出栈 dfs写三个参数:depth搜索深 ...