Sum 判断数组中时候有两个数的和等于一个定值 (数组很大) O(n)

Sum

Time Limit: 9000 MS Memory Limit: 65536 K

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Sum

Mr. Jojer is given n numbers and an extra integer x, he wants to know whether there are two numbers whose sum is x.

Input

The input file contains several test cases. The first line of each test case contains two integers, n(<=1000001) and x. From the next line of each test case, there are n numbers.

Output

Each test case corresponds to a line in the output, which is either "YES" if there exists an answer or "NO" if not.

Sample Input

3 3
1 2 3
2 3
1 3

 

Sample Output

YES
NO

 

 

 

#include<iostream>
#include<algorithm>
using namespace std;
int a[1001],n,goal;
int main()
{
 int i,j;
 while(scanf("%d%d",&n,&goal)==2)
 {
  for(i=0;i<n;i++)
  { 
   scanf("%d",&a[i]);
  }
  sort(a,a+n);
  if(a[n-4]+a[n-1]+a[n-2]+a[n-3]<goal) {printf("NO/n");continue;}
  int flag=0;
  for(i=0;i+3<n;i++)
  {
   if(a[i]+a[i+1]+a[i+2]+a[i+3]>goal) break;
  
   for(j=i+1;j<n;j++)
   {
    int low=j+1,high=n-1,t;
    while (low<high)
    {
     t=a[low]+a[high]+a[i]+a[j];
     if (t==goal) {flag=1;break;}
     else if (t<goal) low++;
     else high--;
    }
    if(flag) break;
   }
   if(flag) break;
  }
  if(flag) printf("YES/n");
  else printf("NO/n");
 }
 return 0;
}