let k=a+b
run in two loops
for(i=0;((i<n) && (a[i]<k));i++)
for(j=i+1;a[j]<=(k-a[i]);j++)
if(a[j]==k-a[i])
do break from both outer n inner loops;
diplay a[i] n a[j]
Time complexity:O(n^2) worst case
shud take O(nlgn) on an average
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