Theory:

**Perfect square problem**.Problem is to determine a pair of numbers where the small (lower_num) number divides the numbers from 1 up to higher_num into two groups in a way that summation of 1 to (lower_num-1) is equal to summation of the numbers from (lower_num+1) to higher_num.

Expanation: first pair 6 8

summation of 1 to 5 = 5*(5+1)/2 = 15

summation of 7 to 8 = 7+8=15

so, 5*(5+1)/2 = 8*(8+1)/2-6-5*(5+1)/2, here n*(n+1)/2 formula is used.

so if we assume the lower_num=divider;

then (divider-1)*divider/2=summation of 1 to higher_num – divider – (divider-1)*divider/2.

=> (divider-1)*divider + divider = summation of 1 to higher_num;

=> divider^2 – divider + divider = Sum_hi;

=> divider = sqrt of Sum_hi;

Solution:

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#include <iostream> #include <cstdio> #include <cmath> #define lld long long using namespace std; int main() { double db; lld a,b,lp,hp; int cnt=1; double i=8; while(cnt<11) { db=sqrt(i*(i+1)/2); a=db; if(a==db) { //cout<<a<<" "<<i<<"\n"; printf("%10ld%10ld\n",a,long(i)); cnt+=1; } i+=1; } return 0; } |