Codeforces Round #272 (Div. 2)-C. Dreamoon and Sums
http://codeforces.com/contest/476/problem/C
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 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}.
解题思路:可推出公式 ans =
1 #include <stdio.h>
2 #include <math.h>
3
4 #define ll long long
5
6 int main(){
7 ll a, b, k, ans, x, y;
8 ll MOD = 1e9 + ;
9 while(scanf("%I64d %I64d", &a, &b) != EOF){
ans = ;
for(k = ; k <= a; k++){
x = (k * b + ) % MOD;
y = (b * (b - ) / ) % MOD;
ans += x * y % MOD;
}
printf("%I64d\n", ans % MOD);
}
return ;
19 }
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