Codeforces Round #272 (Div. 2) C

標籤：algorithm   演算法   acm   codeforces   

題目：C. Dreamoon and Sumstime limit per test1.5 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard 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}.

題意分析：數學題。看懂公式就行了。考慮枚舉一下餘數， 如果m = x%b 則 x = mk*b+m = m(kb+1)  m可以預先處理為所有可能的餘數和，注意long long 和取MOD。
代碼:

#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;    }}

Codeforces Round #272 (Div. 2) C