HDU -- 1868 -- math problems <I have done it all TM>

Mathematical scum hiding =-=

Fortunately, this question is about high school knowledge... the sum of series ..

At the beginning, I thought of a 2-layer method with the largest chest and no brain. Even if I saw this data, I thought it would be <32 bits very big ..

I think about it again. I don't think I can find it in deep search...

Just to simplify the formula.

X1 + X2 + X3 +... + Xn Special tolerances d = 1 so X2 = X1 + 1x3 = X1 + 2 xn = X1 + n-1

Then it can be reduced to X1 * n + N * (n-1) /2 = sum here n is the index Column Length Sum is the question for us to write n letters is not good, forgive me =-= N * (n-1)/2 is 0, 1, 2, 3, 4 .... the sum of N

OK. I thought it was AC.

But I am still tle... I am about it ..

 1 #include <iostream> 2 using namespace std; 3  4 int main() 5 { 6     cin.sync_with_stdio(false); 7     int n , ans , temp; 8     while( cin >> n ) 9     {10         ans = 0;11         for( int i = 2 ; i<=n ; i++ )12         {13             temp = i*(i-1)/2;14             if(n>temp)15             {16                 if( (n-temp)%i == 0 )17                     ans ++;18             }                    19         }20         cout << ans << endl;21     }22     return 0;23 }

In fact, you can also see that I have another judgment on if (n> temp) in for, because some of them will be larger than N, and then subtract is negative, but the same is true (n-Temp) % I = 0 so I need to cut it

However, if I cut it like this, the for side will continue to run.

In fact, it is very simple. I just need to add an else break;

 1 #include <iostream> 2 using namespace std; 3  4 int main() 5 { 6     cin.sync_with_stdio(false); 7     int n , ans , temp; 8     while( cin >> n ) 9     {10         ans = 0;11         for( int i = 2 ; i<=n ; i++ )12         {13             temp = i*(i-1)/2;14             if(n>temp)15             {16                 if( (n-temp)%i == 0 )17                     ans ++;18             }    19             else20                 break;                21         }22         cout << ans << endl;23     }24     return 0;25 }

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