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  • Codeforces Round #272 (Div. 2)-C. Dreamoon and Sums

    http://codeforces.com/contest/476/problem/C

    C. Dreamoon and Sums
    time limit per test
    1.5 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    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?

    Input

    The single line of the input contains two integers ab (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}.

    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 + 7;
     9     while(scanf("%I64d %I64d", &a, &b) != EOF){
    10         ans = 0;
    11         for(k = 1; k <= a; k++){
    12             x = (k * b + 1) % MOD;
    13             y = (b * (b - 1) / 2) % MOD;
    14             ans += x * y % MOD;
    15         }
    16         printf("%I64d ", ans % MOD);
    17     }
    18     return 0;

    19 } 

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  • 原文地址:https://www.cnblogs.com/angle-qqs/p/4027164.html
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