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 15Sample Output 1:
4 11Sample Input 2:
7 14 1 8 7 2 4 11 15Sample Output 2:
No Solution
题意：在一堆数据中找出两个数使得这两个数相加等于给出的数，如果有多种选择输出那个含有最小值的组合。
思路：因为数据量较大，所以暴力搜索超时是肯定的，因此选用折半查找。先排序枚举第一个数字，再查找第二个数字。
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
int arry[100005];
bool cmp(int x,int y)
{
return x<y;
}
bool binarysearch(int key,int l,int r)
{
int left=l,right=r,mid;
while(left<=right)
{
mid=(left+right)/2;
if(key==arry[mid]) return true;
else if(key>arry[mid]) left=mid+1;
else right=mid-1;
}
return false;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
int i,j,k;
for(i=0;i<n;i++)
{
scanf("%d",&arry[i]);
}
sort(arry,arry+n,cmp);
int tag=0,key;
for(i=0;i<n;i++)
{
key=m-arry[i];
if(binarysearch(key,i+1,n-1))
{
tag=1;
j=arry[i];
k=key;
break;
}
}
if(tag) printf("%d %d",j,k);
else printf("No Solution\n");
return 0;
}[折半查找]PAT1048 Find Coins,布布扣,bubuko.com
原文：http://blog.csdn.net/zju_ziqin/article/details/20202037