C. Dreamoon and Sums(Codeforces Round #272)
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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?
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}.
直接推公式就行了。
代码:
//31ms#include <iostream>#include <cstdio>#include <algorithm>using namespace std;const int mod=1000000007;int main(){ long long a,b; scanf("%I64d%I64d",&a,&b); long long ans=0; ans=(b*(b-1)/2%mod)*(((((a+1)*a/2)%mod)*b%mod+a)%mod)%mod; cout<<ans<<endl; return 0;}
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