cf(#div1 A. Dreamoon and Sums)(数论)
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}.
题意:
给你两个整数a,b ,定义这样一个数为漂亮数:
(1) x>0 ; (2)x%b!=0 ; (3) (x/b)/(x%b)=k; (4)k属于[1,a];
求这样的漂亮数的所有之和。
对于这样题,分析起来其实还是很简单的, 对于求和(sum),我们需要做的第一件事就是确定漂亮数的边界(上限和下限)、由于(3) (x/b)/(x%b)=k
可以得出: x/b = k*(x%b); x%b的剩余系即{1,2,3,4,5,......b-1}里的最大系;我们不放设定y={1,2,3,4,5,........b-1};
不难得出: x=k*y*b+y; -->x=(k*b+1)*y 所以可以推断公式: x=Σ1a (k*b+1)*Σ1b-1 (y);
进一步简化后部分: x=Σ1a (k*b+1)*(b-1)*(b)/2;
所以代码为:
#include<cstdio>
#include<cstring>
#define LL __int64
#define mod 1000000007
int main()
{
LL a,b;
LL sum;
while(scanf("%I64d%I64d",&a,&b)!=EOF)
{
sum=;
for(LL j=;j<=a;j++)
{
sum+=((j*b)%mod+)%mod;
sum%=mod;
}
LL ans=(((b-)*b)/)%mod;
printf("%I64d\n",(sum*ans)%mod);
}
return ;
}
cf(#div1 A. Dreamoon and Sums)(数论)的更多相关文章
- cf(#div1 B. Dreamoon and Sets)(数论)
B. Dreamoon and Sets time limit per test 1 second memory limit per test 256 megabytes input standard ...
- codeforces 477A A. Dreamoon and Sums(数学)
题目链接: A. Dreamoon and Sums time limit per test 1.5 seconds memory limit per test 256 megabytes input ...
- Codeforces Round #272 (Div. 2) C. Dreamoon and Sums 数学
C. Dreamoon and Sums time limit per test 1.5 seconds memory limit per test 256 megabytes input stand ...
- 【CODEFORCES】 A. Dreamoon and Sums
A. Dreamoon and Sums time limit per test 1.5 seconds memory limit per test 256 megabytes input stand ...
- CF 984C Finite or not? (数论)
CF 984C Finite or not? (数论) 给定T(T<=1e5)组数据,每组数据给出十进制表示下的整数p,q,b,求问p/q在b进制意义下是否是有限小数. 首先我们先把p/q约分一 ...
- 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 mem ...
- Codeforces Round #272 (Div. 2)C. Dreamoon and Sums 数学推公式
C. Dreamoon and Sums Dreamoon loves summing up something for no reason. One day he obtains two int ...
- Codeforces Round #272 (Div. 1) A. Dreamoon and Sums(数论)
题目链接 Dreamoon loves summing up something for no reason. One day he obtains two integers a and b occa ...
- cf 450b 矩阵快速幂(数论取模 一大坑点啊)
Jzzhu has invented a kind of sequences, they meet the following property: You are given x and y, ple ...
随机推荐
- 关于float的感悟
给元素设置了float样式后,最终的结果是: 1:这个元素漂浮起来, 2:其他的元素位置可以视为 这个元素不存在 的时候的位置:但是float样式还是对整个页面有所影响 3:float的影响就是他附近 ...
- lncRNA研究
------------------------------- Long noncoding RNAs are rarely translated in two human cell lines. ( ...
- hdu 3033 I love sneakers! 分组背包
I love sneakers! Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) ...
- Android内存泄露测试
Android性能测试过程中的一些常用命令: CPU: adb shell top -n | grep "+PackageName 内存: adb shell dumpsys meminfo ...
- Codeforces Round #198 (Div. 2)
A.The Wall 题意:两个人粉刷墙壁,甲从粉刷标号为x,2x,3x...的小块乙粉刷标号为y,2y,3y...的小块问在某个区间内被重复粉刷的小块的个数. 分析:求出x和y的最小公倍数,然后做一 ...
- iOS - Swift Enum 枚举
1.Enum 的创建 1.1 标准定义 枚举的定义 enum CompassPoint { case North case South case East case West } enum Plane ...
- PHP面向对象(OOP)编程入门教程————如何实例化对象?
我们上面说过面向对象程序的单位就是对象,但对象又是通过类的实例化出来的,既然我们类会声明了,下一步就是实例化对象了. 当定义好类后,我们使用new关键字来生成一个对象. $对象名称 = new 类名称 ...
- Can of Worms 【迭代/线段树】
题意:一条直线上有n个炸弹,给出每个炸弹的爆炸半径,可以引爆另一个炸弹爆炸.问:每个炸弹爆炸后,最多有几个炸弹一起爆炸? 迭代,用线段树更新. #include <cstdio> #inc ...
- Vsftpd服务的搭建
安装vsftpd服务程序 yum install vsftpd -y Vsftpd的程序与配置文件: 主程序 /usr/sbin/vsftpd 用户禁止登陆列表 /etc/vsftpd/ftpuser ...
- ssh免密码登陆及其原理
ssh 无密码登录要使用公钥与私钥.linux下可以用用ssh-keygen生成公钥/私钥对,下面我以CentOS为例. 有机器A(192.168.1.155),B(192.168.1.181).现想 ...