【最简解法】1048 Find Coins (25 分)

立志用最少的代码做最高效的表达

PAT甲级最优题解——>传送门

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

题意：输入n，value，接下来输入n个数，要求用其中的两个数相加等于value。如果有多组，则输出最小的。

算法设计：有双指针法和桶排法两种方法。双指针法的时间复杂度是O(nlogn)。桶排法的时间复杂度为O(n)，算法逻辑请阅读代码体会。

代码一：双指针法

#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;
}

代码二：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;
}

耗时(桶排)：

求赞哦~ (✪ω✪)

【最简解法】1048 Find Coins (25 分)_18行代码AC相关推荐

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【最简解法】1048 Find Coins (25 分)_18行代码AC

立志用最少的代码做最高效的表达

代码一：双指针法

代码二：桶排法

耗时(桶排)：

求赞哦~ (✪ω✪)

【最简解法】1048 Find Coins (25 分)_18行代码AC相关推荐

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