Codeforces Round #272 (Div. 1) A. Dreamoon and Sums(数论)
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).
题意 : 给你a,b。让你找出符合以下条件的x,div(x,b)/mod(x,b)=k,其中k所在范围是[1,a],其中mod(x,b)!= 0.然后将所有符合条件的x加和,求最后的结果
官方题解 :
If we fix the value of k, and let d = div(x, b), m = mod(x, b), we have :
d = mk
x = db + m
So we have x = mkb + m = (kb + 1) * m.
And we know m would be in range [0, b - 1] because it's a remainder, so the sum of x of that fixed k would be .
Next we should notice that if an integer x is nice it can only be nice for a single particular k because a given x uniquely definesdiv(x, b) and mod(x, b).
Thus the final answer would be sum up for all individual k: which can be calculated in O(a) and will pass the time limit of 1.5 seconds.
Also the formula above can be expanded to .
#include <stdio.h>
#include <string.h>
#include <iostream> using namespace std ;
#define mod 1000000007 int main()
{
long long a,b ;
while(~scanf("%I64d %I64d",&a,&b)){
// printf("%I64d\n",a*(a+1)/2) ;
long long sum = (((a*(a+)/%mod)*b%mod+a)%mod*(b*(b-)/%mod))%mod ;
printf("%I64d\n",sum) ;
}
return ;
}
Codeforces Round #272 (Div. 1) A. Dreamoon and Sums(数论)的更多相关文章
- 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. 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 Round #272 (Div. 2) C. Dreamoon and Sums (数学 思维)
题目链接 这个题取模的时候挺坑的!!! 题意:div(x , b) / mod(x , b) = k( 1 <= k <= a).求x的和 分析: 我们知道mod(x % b)的取值范围为 ...
- Codeforces Round #272 (Div. 2) E. Dreamoon and Strings 动态规划
E. Dreamoon and Strings 题目连接: http://www.codeforces.com/contest/476/problem/E Description Dreamoon h ...
- Codeforces Round #272 (Div. 2) D. Dreamoon and Sets 构造
D. Dreamoon and Sets 题目连接: http://www.codeforces.com/contest/476/problem/D Description Dreamoon like ...
- Codeforces Round #272 (Div. 2) B. Dreamoon and WiFi dp
B. Dreamoon and WiFi 题目连接: http://www.codeforces.com/contest/476/problem/B Description Dreamoon is s ...
- Codeforces Round #272 (Div. 2) A. Dreamoon and Stairs 水题
A. Dreamoon and Stairs 题目连接: http://www.codeforces.com/contest/476/problem/A Description Dreamoon wa ...
- Codeforces Round #272 (Div. 2) E. Dreamoon and Strings dp
题目链接: http://www.codeforces.com/contest/476/problem/E E. Dreamoon and Strings time limit per test 1 ...
随机推荐
- [转]Script标签和脚本执行顺序
Script标签和脚本执行顺序 这里详细聊聊和script标签相关的脚本执行顺序. Script标签的默认行为 几个首要特性: script标签(不带defer或async属性)的会阻止文档渲染.相关 ...
- js判断客户端是pc还是手机
function IsPC() { var userAgentInfo = navigator.userAgent; var Agents = ["Android", " ...
- bzoj 4556 字符串 —— 后缀数组+主席树
题目:https://www.lydsy.com/JudgeOnline/problem.php?id=4556 就是找一个 rk 在一段区间内的前驱和后继: 由于 LCP 还有区间长度的限制,所以可 ...
- DLatch by Verilog
//-----------------------------------------------------// Design Name : dlatch_reset// File Name : ...
- 在actionbar中加入item的方法
首先在menu文件夹中创建post.xml <?xml version="1.0" encoding="utf-8"?> <menu xmln ...
- android生命周期参考
public class ActivityDemo extends Activity { private static final String TAG = "ActivityDemo&qu ...
- ROS探索总结(三)——ROS新手教程
转自古-月 http://blog.csdn.net/hcx25909 前面我们介绍了ROS的特点和结构,接下来就要开始准备动手感受一下ROS的强大了. ROS官网的wiki上针对新手的教程很详细,最 ...
- 如何用FFmpeg API采集摄像头视频和麦克风音频,并实现录制文件的功能
之前一直用Directshow技术采集摄像头数据,但是觉得涉及的细节比较多,要开发者比较了解Directshow的框架知识,学习起来有一点点难度.最近发现很多人问怎么用FFmpeg采集摄像头图像,事实 ...
- 【转】Jmeter做功能测试的优点和不足
利用Jmeter做功能测试有以下优点: ● 不依赖于界面,如果服务正常启动,传递参数明确就可以添加测试用例,执行测试 ● 测试脚本不需要编程,熟悉http请求,熟悉业务流程,就可以根据页面 ...
- java求10!的阶乘
package com.aaa; //求10!的阶乘 public class Cheng { public static void main(String[] args) { int s=1; fo ...