1048 Find Coins（25 分）

1048 Find Coins 二分查找解题

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博客围绕 1048 Find Coins 题目展开，该题要求从众多硬币中找出两枚，使其面值之和等于给定金额。输入包含硬币总数和需支付金额，以及各硬币面值。输出满足条件的两枚硬币面值，若无解则输出 No Solution，解题思路为二分查找。

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 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

解题思路：
二分查找。

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=1e5+10;
int a[maxn];
bool cmp(int a,int b){
    return a<b;
}
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    sort(a,a+n,cmp);
    int i=0,j=n-1;
    while(i<j){
        if(a[i]+a[j]==m){
            printf("%d %d",a[i],a[j]);
            return 0;
        }else if(a[i]+a[j]<m){
            i++;
        }else{
            j--;
        }
    }
    printf("No Solution\n");
    return 0;
}