Codeforces round #272 (Div. 1) A. dreamoon and sums (number theory)

Question Link

Dreamoon loves summing up something for no reason. One day he obtains two integersAAndBOccasionally. He wants to calculate the sum of all nice integers. Positive IntegerXIs called nice if and, whereKIs some integer number in range [1, random,A].

By we denote the quotient of integer divisionXAndY. By we denote the remainder of integer divisionXAndY. You can read more about these operations here: http://goo.gl/AcsXhT.

The answer may be large, so please print its remainder modulo 1 limit 000 limit 000 limit 007 (10^{9 rows + 7). Can you compute it faster than dreamoon?}

Input

The single line of the input contains two integersA,B(1 digit ≤ DigitA, Bytes,BLimit ≤ limit 10^{7 ).}

Output

Print a single integer representing the answer modulo 1 0000000 0000000 0000007 (10^{9 rows + rows 7 ).}

 

Question: A and B. Let you find the X, Div (X, B)/MoD (X, B) = k that meets the following conditions, where k is in the range of [1, A], where Mod (X, B )! = 0. Then sum all X that meets the condition and calculate the final result.

Official question:

If we fix the value K , And let D Signature = Signature Div ( X , Bytes, B ),M Signature = Signature MoD ( X , Bytes, B ), We have:
D Signature = Signature Mk
X Signature = Signature DB Region + Region M
So we have X Signature = Signature MKB Region + Region M Signature = Signature ( KB Rows + rows 1) rows * Rows )* M .
And we know M Wocould be in range [0, random, B Using-keys 1] because it's a remainder, so the sum X Of that fixed K Wocould be.
Next we shoshould notice that if an integer X Is Nice It can only be Nice For a single particle K Because a given X Uniquely defines Div ( X , Bytes, B ) And MoD ( X , Bytes, B ).
Thus the final answer wocould be sum up for all individual K : Which can be calculated in O ( A ) And will pass the time limit of 1.5 seconds.
Also the formula abve can be expanded.

#include <stdio.h>#include <string.h>#include <iostream>using namespace std ;#define mod 1000000007int main(){    long long a,b ;    while(~scanf("%I64d %I64d",&a,&b)){        //    printf("%I64d\n",a*(a+1)/2) ;        long long sum = (((a*(a+1)/2%mod)*b%mod+a)%mod*(b*(b-1)/2%mod))%mod ;        printf("%I64d\n",sum) ;    }    return 0 ;}

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Codeforces round #272 (Div. 1) A. dreamoon and sums (number theory)