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 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 = M and 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
 #include <stdio.h>
int main()
{
int n,line,i;
int hash[];
while(scanf("%d%d",&n,&line)!=EOF)
{
getchar();
for(i = ;i<=;i++)
hash[i] = ;
for(i = ;i<n;i++)
{
int tem;
scanf("%d",&tem);
++hash[tem];
} bool find = false;
for(i = ;i<= line/;++i)
{
if(hash[i]>)
{
--hash[i];
if(hash[line-i]>)
{
find = true;
break;
}
else
{
++hash[i];
}
}
} if(find)
{
printf("%d %d\n",i,line-i);
}
else
{
printf("No Solution\n");
} }
return ;
}

1048. Find Coins (25)的更多相关文章

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 1048 Find Coins (25分)

    Eva loves to collect coins from all over the universe, including some other planets like Mars. One d ...

  7. PAT (Advanced Level) 1048. Find Coins (25)

    先对序列排序,然后枚举较小值,二分较大值. #include<iostream> #include<cstring> #include<cmath> #includ ...

  8. PAT甲题题解-1048. Find Coins (25)-水

    给n,m以及n个硬币 问,是否存在两个硬币面值v1+v2=m 因为面值不会超过500,所以实际上最多500个不同的硬币而已 #include <iostream> #include < ...

  9. 【PAT甲级】1048 Find Coins (25 分)(二分)

    题意: 输入两个正整数N和M(N<=10000,M<=1000),然后输入N个正整数(<=500),输出两个数字和恰好等于M的两个数(小的数字尽可能小且输出在前),如果没有输出&qu ...

随机推荐

  1. poj2299解题报告(归并排序求逆序数)

    POJ 2299,题目链接http://poj.org/problem?id=2299 题意: 给出长度为n的序列,每次只能交换相邻的两个元素,问至少要交换几次才使得该序列为递增序列. 思路: 其实就 ...

  2. css实现居中

    --在html常常用到居中 --1.可以用<center></center> --2.可以用css 演示代码: <!DOCTYPE html PUBLIC "- ...

  3. SQL in与exists相关性能问题总结

    SQL  in与exists相关性能问题总结 in 和 exists in 和 exists的是DBA或开发人员日常工作学习中常用的基本运算符,今天我就这两个所带来的性能问题进行分析总结,方便自己与他 ...

  4. Class类中getMethods() 与getDeclaredMethods() 方法的区别

    一:jdk API中关于两个方法的解释 1:getMethods(),该方法是获取本类以及父类或者父接口中所有的公共方法(public修饰符修饰的) 2:getDeclaredMethods(),该方 ...

  5. [改善Java代码]正确使用String,StringBuffer,StringBuilder

    CharSequence接口有三个实现类与字符串有关:String,StringBuffer,StringBuffer.虽然它们都与字符串有关,但是其处理机制是不同的. String类是不可改变的量, ...

  6. KMP模版

    #include<iostream> #include<cstdio> #include<cstring> using namespace std; ]; void ...

  7. 使用git ftp发布我个人的hexo博客内容

    自己虚拟主机中的博客是由hexo3 + next主题,因为我想将 hexo 编译生成的文件可以通过ftp命令发布到ftp服务器上面. 发布使用的工具是git-ftp: 按照Use Jenkins an ...

  8. Dynamic\Static\IsKinematic

    1.Dynamic: 有Collider和RigidBody的GameObject, Unity视之为Dynamic. 适用于经常变换移动的对象. 2.Static: 只含有Collider的Game ...

  9. framework7学习笔记

    最近因项目需要学习framework7,简称F7.对于自己遇到的问题和学习到的东西做个简单记录. 问题:刚开始获取json,页面上一直不显示,不得其法,原来是json文件需要在弹出层打开之后在来加载, ...

  10. document.ready和onload的区别

    转自:http://holysonll.blog.163.com/blog/static/2141390932013411112823855/ document.ready和onload的区别——Ja ...