PAT-A1048-Find Coins

PAT Find Coins - 題目連結

題目描述

 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 e5coins 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.

輸入描述

 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.

輸出描述

 For each test case, print in one line the two face values V 1​ and V 2​ (separated by a space) such that V 1 +V 2​ =M and V 1​ ≤V 2​. If such a solution is not unique, output the one with the smallest V 1​ . If there is no solution, output No Solution instead.

示例：

輸入

8 15

1 2 8 7 2 4 11 15

輸出

4 11

輸入

7 14

1 8 7 2 4 11 15

輸出

No Solution

解題思路

 使用數組存儲每個數字出現的次數，題目需要注意的是當出現兩個想等的數相加時，需要判斷該數字的次數是否大于等于2，不然就會出現錯誤。

參考代碼

#include <iostream>

using namespace std;
int HashTable[1010] = {0};

int main() {
    int n, k, a;
    cin >> n >> k;
    for (int i = 0; i < n; ++i) {
        cin >> a;
        HashTable[a]++;
    }
    for (int i = 1; i < k; ++i) { //隻讀取循環到k-1因子 進行優化
        if (HashTable[i] && HashTable[k - i]) {
            if (i == k - i && HashTable[i] <= 1) { //這裡當被加數和加數相等時，要特殊判斷
                continue;
            }
            printf("%d %d\n", i, k - i);
            return 0;
        }
    }
    printf("No Solution\n");
    return 0;
}