题目：
C. Dreamoon and Sums
time limit per test
1.5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Dreamoon loves summing up something for no reason. One day he obtains two integers a and boccasionally.
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 theremainder 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 test(s)
input
1 1
output
0
input
2 2
output
8
Note
For the first sample, there are no nice integers because 
is
always zero.
For the second sample, the set of nice integers is {3,?5}.
题意分析：
数学题。看懂公式就行了。考虑枚举一下余数， 如果m = x%b 则 x = mk*b+m = m(kb+1) m可以预处理为所有可能的余数和，注意long long 和取MOD。
代码:
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <iostream>
using namespace std;
const int MOD=1000000007;
int main()
{
long long a,b;
while(cin>>a>>b)
{
long long m=(1+b-1)*(b-1)/2%MOD;
//cout<<m<<endl;
long long sum=0;
{
for(long long i=1;i<=a;i++)
{
sum=(sum%MOD+m*((b*i)%MOD+1)%MOD)%MOD;
}
}
cout<<sum<<endl;
}
}