CodeForce-476C-Dreamoon and Sums

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原题：
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 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).

Examples
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：
给定两个数值a，b。现在有一个x满足x与b的商除以x与b的余数得到的数值记为k，k要求在1到a之间（包括1和a）现在要求你求出所有的x的和，并且结果和1000000007取余。
题解：
这道题目，逆向思维思考一下，反着来求出所有的x，即通过遍历所有的a和b求出所有的x：

        for(int i=1;i<b;i++)//a、b遍历顺序随意
        {
            for(int j=1;j<=a;j++)
            {
                long long res=i*j*b+i;//res=(j*b+1)*i
                sum+=res;
                sum=sum%1000000007;
            }
        }

但是光是这样，那坑定会超时的。所以我们可以通过提取公因式，这里可以提出i，利用等差数列求和即可。
但是这里要注意数据范围，a^3 的范围大约是10的21次方，超过了longlong的范围。
附上AC代码：

#include<cstdio>
using namespace std;
#define LL long long
LL ans = 0;
const LL mod = 1e9+7;
int main()
{
   LL a,b;
   scanf("%lld%lld",&a,&b);
   LL m = (b-1)*b/2;
   m%=mod;
   for(int i = 1;i<=a;i++)
    {
		LL tmp = (i*b+1)%mod;//无时无刻不取余，降低数据范围
		ans = (ans + m*tmp%mod)%mod;
	}
   printf("%lld\n",ans);
}

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