PAT甲级 1048 Find Coins (25point(s))-优快云博客

本文链接：https://blog.youkuaiyun.com/Seqlist_base/article/details/104577173

面对宇宙购物商场的特殊支付要求，Evalove需要从成千上万的硬币中找到恰好两枚，其面值之和等于特定金额。此任务探讨了算法优化与数据结构应用，特别是在大量数据中快速查找配对的问题。

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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 10^5 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 (≤10^5
, the total number of coins) and M (≤10^3, 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 V1and 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
这道题如果用暴力双for循环会有两个节点超时，还有就是注意当v[i]=m-v[i]时的特殊情况（不然会有一个点过不了）；

#include<cstdio>
#include<algorithm>
#include<vector>
#include<iostream>
using namespace std;
int a[100001]={0};
vector<int>v;
int main(){
int n,m,temp;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++){
    scanf("%d",&temp);
    v.push_back(temp);
    a[temp]++;
}
sort(v.begin(),v.end());
for(int i=0;i<n-1;i++){
if(a[m-v[i]]!=0){
    if(((m-v[i])==v[i])&&(a[v[i]]<2)){
continue;
    }
    printf("%d %d",v[i],m-v[i]);
return 0;
    }
}
 printf("No Solution");
system("pause");
return 0;
}