题目

解法1

点击查看代码 
#include <iostream>
#include <unordered_map>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n, m;
    cin >> n >> m;
    vector<int> coins(n);
    unordered_map<int, int> countMap;

    for (int i = 0; i < n; ++i) {
        cin >> coins[i];
        countMap[coins[i]]++;
    }

    int v1 = -1, v2 = -1;
    sort(coins.begin(), coins.end());  // 确保找到的解是 V1 最小的

    for (int i = 0; i < n; ++i) {
        int a = coins[i];
        int b = m - a;
        if (b < a) continue; // 确保 V1 <= V2，且不重复输出

        // 减去当前 a 的计数临时查找 b
        countMap[a]--;
        if (countMap[b] > 0) {
            v1 = a;
            v2 = b;
            break;
        }
        countMap[a]++;
    }

    if (v1 == -1) {
        cout << "No Solution" << endl;
    } else {
        cout << v1 << " " << v2 << endl;
    }

    return 0;
}

解法2：二分查找

点击查看代码 
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
 
int main(){
    int N, M, i, j;
    scanf("%d %d", &N, &M);
    
    vector<int> vec(N);
    for(i = 0; i < N; ++i){
        scanf("%d", &vec[i]);
    }
    
    sort(vec.begin(), vec.end());
    
    i = 0;
    j = vec.size() - 1;
    while(i < j){
        if(vec[i] + vec[j] == M){
            printf("%d %d", vec[i], vec[j]);
            return 0;
        }
        if(vec[i] + vec[j] < M){
            i++;
        } else{
            j--;
        }
    }
    
    printf("No Solution");
    return 0;
}