1048. Find Coins (25)

题目：

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:

8 151 2 8 7 2 4 11 15

Sample Output 1:

4 11

Sample Input 2:

7 141 8 7 2 4 11 15

Sample Output 2:

No Solution

注意：

1、这里要使用二分法，不然会超时。

2、思路是先将所有的coin排序，然后依次二分查找有没有匹配的另一个coin。

3、注意停止条件，当需要匹配的那个coin已经是pay的一半的时候就可以停止循环了，因为后面已经全部是大于这个coin值的coin了，加起来肯定全部大于pay的。

代码：

//1048#include<iostream>#include<vector>#include<algorithm>using namespace std;int main(){int n,m;vector<int>money;scanf("%d%d",&n,&m);for(int i=0;i<n;++i){int k;scanf("%d",&k);if(k<m)//erase the element which is greater than paymentmoney.push_back(k);}if(money.size()==0){printf("No Solution");return 0;}sort(money.begin(),money.end());for(int i=0;i<money.size()-1 && money[i]<=m/2;++i){int start=i+1,end=money.size()-1;while(end>start){//binary searchint mid=(start+end)/2;if(money[i]+money[mid]>m)end=mid;else if(money[i]+money[mid]<m)start=mid+1;else break;}if(money[i]+money[start]==m){printf("%d %d",money[i],money[start]);return 0;}}printf("No Solution");return 0;}