#include<stdio.h>

int main(void){
    long n,m,i,j;
    int flag;
    int A[50001],B[50001];
    scanf("%ld",&n);
    for(i=1;i<=n;i++)
        scanf("%d",&A[i]);
    scanf("%ld",&m);
    for(j=1;j<=m;j++)
        scanf("%d",&B[j]);
    flag=0;
    for(i=1;i<=n;i++)
        for(j=1;j<=m;j++)
            if(A[i]+B[j]==10000)
                flag=1;
    if(flag)
        printf("YES");
    else
        printf("NO");
return 0;
}

not an optimal algo... O(n^2), this problem can be solved with O(n+m)
you should think a little more.

If you need code, send an email on andrewshmig@gmail.com

Edited by author 06.05.2012 00:27

or u can use BinarySearch to search second value.

for example in the first test:
1) u take -175 value,
2) try to find 10175 in second list (using binary search) (because -175 + 10175 == 10000)
...
good luck!

Edited by author 08.05.2012 00:20

>not an optimal algo... O(n^2), this problem can be solved with O(n+m)

This problem can be solved with O(n). )

Edited by author 02.06.2012 20:52

Use the associative containers

you can use HashSet, it's faster

use the two-pointers method or boolean set. It's enough fast for this problem.