Codeforces Round #272 (Div. 2) C. Dreamoon and Sums 数学
1.5 seconds
256 megabytes
standard input
standard output
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).
1 1
0
2 2
8
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}.
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e5+,M=1e6+,inf=1e9+,mod=1e9+;
int main()
{
ll a,b;
scanf("%lld%lld",&a,&b);
ll ans=;
for(ll i=;i<b;i++)
{
ans=ans+(a*((i*b+i)%mod))%mod+(((b*i)%mod)*((a*(a- )/)%mod))%mod;
ans%=mod;
}
printf("%lld\n",ans);
return ;
}
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