要想到利用a到b的求和公式
#include<iostream>
using namespace std
;
int main()
{
int n
;
cin
>> n
;
for (int i
= 1; i
<= n
/ 2; i
++)
{
for (int j
= i
+ 1; j
<= n
/ 2; j
++)
{
if ((i
+ j
)*(j
- i
+ 1) / 2 == n
)
cout
<< i
<< " " << j
<< endl
;
}
}
return 0;
}
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