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 
and 
, where k is some integer number in range[1, a].
By 
we denote the quotient of integer division of x and y. By 
we denote the remainder of integer division of x andy. 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?
The single line of the input contains two integers a, b (1 ≤ a, b ≤ 107).
Print a single integer representing the answer modulo 1 000 000 007 (109 + 7).
题意 ： 给你a，b。让你找出符合以下条件的x，div(x,b)/mod(x,b)=k，其中k所在范围是[1,a]，其中mod(x,b)!= 0.然后将所有符合条件的x加和，求最后的结果
官方题解 ：
If we fix the value of k, and let d = div(x, b), m = mod(x, b), we have :
d = mk
x = db + m
So we have x = mkb + m = (kb + 1) * m.
And we know m would be in range [0, b - 1] because it's a remainder, so the sum of x of that fixed k would be 
.
Next we should notice that if an integer x is nice it can only be nice for a single particular k because a given x uniquely definesdiv(x, b) and mod(x, b).
Thus the final answer would 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 above can be expanded to 
.
#include <stdio.h> #include <string.h> #include <iostream> using namespace std ; #define mod 1000000007 int main() { long long a,b ; while(~scanf("%I64d %I64d",&a,&b)){ // printf("%I64d ",a*(a+1)/2) ; long long sum = (((a*(a+1)/2%mod)*b%mod+a)%mod*(b*(b-1)/2%mod))%mod ; printf("%I64d ",sum) ; } return 0 ; }