1048 Find Coins (25 分)

最新推荐文章于 2024-03-31 15:52:13 发布

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最新推荐文章于 2024-03-31 15:52:13 发布

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原文链接：http://www.cnblogs.com/zllwxm123/p/11185190.html

博客围绕Eva用硬币支付的问题展开，支付要求是用两枚硬币恰好支付指定金额。介绍了输入规范，包含硬币总数和需支付金额等信息；输出规范为找出满足条件的两枚硬币面值，若有多种解则输出最小V1的组合，无解则输出特定内容，还给出了示例输入输出。

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1048 Find Coins (25 分)

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 $1 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.$

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers:

Output Specification:

For each test case, print in one line the two face values

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


看到500就好办了，不要去看另一个，不然可能会卡时间（我没试，猜的）

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int n, m;
 4 int an[100005];
 5 int vis[1000];
 6 int main(){
 7     scanf("%d%d", &n, &m);
 8     for(int i = 0; i < n; i++){
 9         scanf("%d", &an[i]);
10         vis[an[i]]++;
11     }
12 
13     bool flag = true;
14     for(int i = 0; i <= 500; i++){
15         if(vis[i]){
16             vis[i]--;
17             if(vis[m-i]){
18                 printf("%d %d\n", i, m-i);
19                 flag = false;
20                 break;
21             }
22             vis[i]++;
23         }
24     }
25     if(flag){
26         printf("No Solution\n");
27     }
28     return 0;
29 }