PAT 甲級 1048.Find Coins C++/Java

題目來源

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

題意：

給出兩個數N和M

接下來輸入N個數字

是否存在兩個數 a 和 b，知足 a + b = M

• 存在：依次輸出a和b（知足a ≤ b）
• 不存在：輸出 No Solution

分析：

1. 雙指針

先對數組排序，初始化兩個指針left和right，分別指向數組兩端

• left + right < m：left++
• left + right > m：right--
• left + right == m：輸出 left right
• 當left <= right 退出循環

2. 哈希表

出現過的數字存入表中，記錄出現過的次數

在hash中找到一對數字 hash[i] 和 hash[m - i]，它們的和爲 m

注意：

• 當 i = m - i，必須保證數字 i 的個數大於等於2，不然是不能夠的
  • 好比須要14，有兩個7，知足 7 + 7 = 14

c++實現：

1. 雙指針

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    int n;  // the total number of coins
    int m;  // the amount of money Eva has to pay
    cin >> n >> m;
    vector<int> arr(n);
    for (int i = 0; i < n; ++i) {
        cin >> arr[i];
    }
    sort(arr.begin(), arr.end());
    int left = 0;
    int right = n - 1;
    while (left != right) {
        if (arr[left] + arr[right] > m) {
            right--;
        }
        else if (arr[left] + arr[right] < m) {
            left++;
        }
        else {
            cout << arr[left] << ' ' << arr[right];
            return 0;
        }
    }
    cout << "No Solution";
    return 0;
}

2. 哈希表

#include <iostream>
using namespace std;

int arr[1001] = { 0 };
int main() {
    int n;  // the total number of coins
    int m;  // the amount of money Eva has to pay
    cin >> n >> m;
    int temp;
    for (int i = 0; i < n; ++i) {
        cin >> temp;
        arr[temp]++;
    }
    for (int i = 0; i < 1001; ++i) {
        if (arr[i] && arr[m - i]) {
            if (i == m - i && arr[i] <= 1) {
                continue;
            }
            cout << i << ' ' << m - i;
            return 0;
        }
    }
    cout << "No Solution";
    return 0;
}

Java實現：