Problem B: Handing Out Candies

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 258  Solved: 19
[Submit][Status][Web Board]

Description

After the 40th ACM-ICPC, Diao Yang is thinking about finding a girlfriend because he feels very lonely when doing ACM all the time. But because of his philandering, he finally decided to find N girlfriends. To achieve his goal, he wanted to find one girlfriend every day for N days continue. That is to say, at the ith day, he will have i girlfriends exactly.

In order to make his N girlfriends happy, he decided to buy candies everyday for N days continue. Every day all of his girlfriends can get candies, and he will give each of them the same amount of candies and the amount will be as much as possible. Then if there are some candies left, he will eat them by himself.

Now the problem is, Diao Yang want to know how many candies he can eat total by himself after N days continue.

Input

The first line contains an integer T, indicating the total number of test cases. Each test case is a line with two integers N

15N" style="box-sizing: border-box; width: 9pt; height: 15.75pt;"> and M (

151鈮?/m:t>N&lolt;231" style="box-sizing: border-box; width: 63.75pt; height: 15.75pt;">  , 

150鈮?/m:t>M&lolt;231" style="box-sizing: border-box; width: 65.25pt; height: 15.75pt;"> ).

Output

For each test case, output the answer in one line.

Sample Input

2
5 7
6 4

Sample Output

7
9
    N
题意:求∑ ( m%i);
    i=1
思路:会发现M对一个数取模成等差数列;
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
#define esp 0.00000000001
//#pragma comment(linker, "/STACK:102400000,102400000")
int scan()
{
int res = , ch ;
while( !( ( ch = getchar() ) >= '' && ch <= '' ) )
{
if( ch == EOF ) return << ;
}
res = ch - '' ;
while( ( ch = getchar() ) >= '' && ch <= '' )
res = res * + ( ch - '' ) ;
return res ;
}
int main()
{
ll x,y,z,i,t;
scanf("%lld",&x);
while(x--)
{
scanf("%lld%lld",&y,&z);
ll ans=max(y-z,0LL)*z;
ll huan=min(y,z);
for(i=;i<=huan;i++)
{
if(z%i!=)
{
ll d=z/i;
ll maxx=min((z%i)/d+,huan-i+);
d=-d;
ans+=(z%i)*maxx+(maxx*(maxx-)/)*d;
i+=maxx-;
}
}
printf("%lld\n",ans);
}
return ;
}

华中农业大学预赛 B Handing Out Candies 余数的和的更多相关文章

  1. 2016华中农业大学预赛 B 数学

    Problem B: Handing Out Candies Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 272  Solved: 20[Submit ...

  2. 2016华中农业大学预赛 E 想法题

    Problem E: Balance Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 205  Solved: 64[Submit][Status][We ...

  3. 寄生线虫免疫学研究新路径!华中农业大学胡敏团队报道寄生线虫N-糖基化修饰图谱

    N-糖基化修饰是真核生物中一种重要的蛋白质翻译后修饰,在许多生物学过程中起着关键作用,包括蛋白质折叠.受体-配体相互作用.免疫应答和疾病发病机制等.近年来,高精度质谱技术的出现促进了糖组和糖蛋白质组的 ...

  4. ACM-ICPC 2018 焦作赛区网络预赛- G:Give Candies(费马小定理,快速幂)

    There are N children in kindergarten. Miss Li bought them NNN candies. To make the process more inte ...

  5. ACM-ICPC 2018 焦作赛区网络预赛 G题 Give Candies

    There are NN children in kindergarten. Miss Li bought them NN candies. To make the process more inte ...

  6. [HZAU]华中农业大学第四届程序设计大赛网络同步赛

    听说是邀请赛啊,大概做了做…中午出去吃了个饭回来过掉的I.然后去做作业了…… #include <algorithm> #include <iostream> #include ...

  7. (hzau)华中农业大学第四届程序设计大赛网络同步赛 G: Array C

    题目链接:http://acm.hzau.edu.cn/problem.php?id=18 题意是给你两个长度为n的数组,a数组相当于1到n的物品的数量,b数组相当于物品价值,而真正的价值表示是b[i ...

  8. 华中农业大学新生赛C题

    http://acm.hzau.edu.cn/problem.php?id=1099 题意: 输入两个整数 l 和 n,代表半径和output的保留小数点位数. 输出圆的面积,保留n位小数. 一开始觉 ...

  9. 华中农业大学第五届程序设计大赛网络同步赛-L

    L.Happiness Chicken brother is very happy today, because he attained N pieces of biscuits whose tast ...

随机推荐

  1. pop to 特定的UIViewController

    1. 我们可以推出到特定的UIViewController 2. 有一个类没有navigationController,以前一般用delegate,我觉得我们可以把引用一个navigationCont ...

  2. 机器学习理论基础学习12---MCMC

    作为一种随机采样方法,马尔科夫链蒙特卡罗(Markov Chain Monte Carlo,以下简称MCMC)在机器学习,深度学习以及自然语言处理等领域都有广泛的应用,是很多复杂算法求解的基础.比如分 ...

  3. windows7下docker配置镜像加速

    原文地址:https://blog.csdn.net/slibra_L/article/details/77505003 1,本文目的:windows7下docker配置镜像加速,下面是具体操作步骤: ...

  4. .NET RSA解密、签名、验签

    using System; using System.Collections.Generic; using System.Text; using System.IO; using System.Sec ...

  5. windows下docker的安装并使用

    硬件虚拟化:硬件虚拟化是一种对计算机或操作系统的虚拟.虚拟化对用户隐藏了真实的计算机硬件,表现出另一个抽象计算平台. 打开任务管理器的性能查看是否支持虚拟化技术 下载windows docker ht ...

  6. Django初级手册4-表单与通用视图

    表单的编写 1. detail.html模版的编写 <h1>{{ poll.question }}</h1> {% if error_message %}<p>&l ...

  7. 如何在Linux环境下通过uwgsi部署Python服务

    部署python程序时常常会遇到同一台服务器上2.x和3.x共存的情况,不同应用需要使用不用的python版本,使用virtualenv创建虚拟环境能很好地解决这一问题. 首先,需要在服务器上安装vi ...

  8. bzoj1647 / P1985 [USACO07OPEN]翻转棋

    P1985 [USACO07OPEN]翻转棋 其实我们只要枚举第一行的状态,后面的所有状态都是可以唯一确定的. 用二进制枚举灰常方便 #include<iostream> #include ...

  9. Android 实践项目开发 总结

      Android 实践项目开发 总结 课程:移动平台应用开发实践  班级:201592  姓名:杨凤  学号:20159213 成绩:___________       指导老师:娄嘉鹏       ...

  10. C++ shared_ptr的用法

    一. http://www.cnblogs.com/welkinwalker/archive/2011/10/20/2218804.html 二.http://www.cnblogs.com/Tian ...