Code Forces 476C Dreamoon and Sums

最新推荐文章于 2018-10-28 21:05:20 发布

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本文提供了解决数学竞赛中计算特定范围内符合条件整数和并求模的算法，通过推导公式快速解决此类问题。

D - D

Time Limit:1500MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 476C

Description

Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occasionally. He wants to calculate the sum of all nice integers. Positive integer x is called nice if $\text{[math]}$ and $\text{[math]}$ , where k is some integer number in range [1, a].

By $\text{[math]}$ we denote the quotient of integer division of x and y. By $\text{[math]}$ we denote the remainder of integer division of x and y. You can read more about these operations here: http://goo.gl/AcsXhT.

The answer may be large, so please print its remainder modulo 1 000 000 007 (10⁹ + 7). Can you compute it faster than Dreamoon?

Input

The single line of the input contains two integers a, b (1 ≤ a, b ≤ 10⁷).

Output

Print a single integer representing the answer modulo 1 000 000 007 (10⁹ + 7).

Sample Input

Input

1 1

Output

Input

2 2

Output

Hint

For the first sample, there are no nice integers because $\text{[math]}$ is always zero.

For the second sample, the set of nice integers is {3, 5}.

水题推公式就行，数学题，x%b的值可以从0取到b-1.加起来就是0+1+2+3+......+b-1.等差数列求和。（0+b-1）*b/2=b*(b-1)/2;

#include <stdio.h>
#define MOD 1000000007
int main()
{
        long long a,b;
        scanf("%I64d%I64d",&a,&b);
        long long sum=0;
        if(b!=1)
                for(int k=1;k<=a;k++)
                        sum=(sum+((b*(b-1)/2)%MOD)*((k*b+1)%MOD))%MOD;
        printf("%I64d\n",sum);
}