PTA甲级——1048

1048 Find Coins

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:

1
2
8 15
1 2 8 7 2 4 11 15

Sample Output 1:

1
4 11

Sample Input 2:

1
2
7 14
1 8 7 2 4 11 15

Sample Output 2:

1
No Solution

思路

​ 大致题意：输入n,m,然后输入n个数，需要你找出两个数使得两个数V1和V2，这里的V1<=V2的和为m,如果存在多个答案则输出V1最小的那组数V1 V2。

​ 思路：也可以选择输入的时候我们直接统计各个数出现次数，然后我们再直接枚举1～500查看是否有这么两个数同时存在可以组成m。当然这个思路只有在m比较小的情况下可以用，其他情况的话大概率会超时的。

代码

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#include<iostream>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<string>
#include<cstring>
using namespace std;
int v[1005];//505会出现段错误
int main(){
    int n,m,x;
    memset(v,0,sizeof(v));
    cin>>n>>m;
    for(int i=1;i<=n;i++){
        cin>>x;
        v[x]++;
    }
    int f=0;
    for(int i=1;i<=500;i++){
        if(i!=m-i){
            if(v[i]&&v[m-i]){
                f=1;
                cout<<i<<" "<<m-i;
                break;
            }
        }else if(i==m-i&&v[i]>=2){
                f=1;
                cout<<i<<" "<<m-i;
                break;
        }
    }
    if(!f) cout<<"No Solution";
    return 0;
}