A1048. Find Coins

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

 1 #include<cstdio>
 2 using namespace std;
 3 const int N = 1005;
 4 int hashTB[N] = {0, 0};
 5 int main(){
 6     int n, m, v;
 7     scanf("%d %d", &n, &m);
 8     for(int i = 0; i < n; i++){
 9         scanf("%d", &v);
10         hashTB[v]++;
11     }
12     for(int i = 0; i < N; i++){
13         if(hashTB[i] && hashTB[m - i]){
14             if(i == m - i && hashTB[i] <= 1){
15                 continue;
16             }
17             printf("%d %d
", i, m - i);
18             return 0;
19         }
20     }
21     printf("No Solution
");
22     return 0;
23 }

View Code

总结：

1、本题题意：给出拥有的硬币，给出要凑出的面额M，找出两个硬币且他们的和为M。显然暴力法很可能超时，所以关键是要想到用哈希表。

2、由于每个钱的面额都小于500，但M可能大于500，故hashTB的大小最好为1000，否则用指针i遍历时，M - i可能越界。

3、此题还可以用二分法。先对所有面值进行排序，遍历a[0]、a[1]……,对于每一个a[i]，寻找是否存在M - a[ i ]且 M - a[i] 不是a[i]，若找到则输出。