立志用最少的代碼做最高效的表達

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PAT甲級最優題解——>傳送門

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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 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.

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

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題意：輸入n，value， 接下來輸入n個數， 要求用其中的兩個數相加等于value。 如果有多組，則輸出最小的。

算法設計：有雙指針法和桶排法兩種方法。 雙指針法的時間復雜度是O(nlogn)。 桶排法的時間復雜度為O(n)， 算法邏輯請閱讀代碼體會。

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代碼一：雙指針法

#include<bits/stdc++.h> using namespace std; int ary[100010]; int main() {ios::sync_with_stdio(false);int num, value; cin >> num >> value;for(int i = 0; i < num; i++) cin >> ary[i];sort(ary, ary+num);int sta = 0, fin = num-1;while(sta != fin) {if(ary[sta] + ary[fin] > value) fin--;else if(ary[sta] + ary[fin] < value) sta++;else { cout << ary[sta] << ' ' << ary[fin] << '\n'; return 0; }}cout << "No Solution\n";return 0; }

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代碼二：桶排法

#include<bits/stdc++.h> using namespace std; int ary[1010]; int main() {ios::sync_with_stdio(false);int num, value, x; cin >> num >> value;for(int i = 0; i < num; i++) { cin >> x; ary[x]++; }for(int i = 0; i < 1005; i++) if(ary[i]) {ary[i]--;if(ary[value-i]) { cout << i << ' ' << value-i; return 0;}}cout << "No Solution";return 0; }

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耗時(桶排)：

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求贊哦~ (?ω?)

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