A. Johny Likes Numbers
time limit per test

0.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Johny likes numbers n and k very much. Now Johny wants to find the smallest integer x greater than n, so it is divisible by the number k.

Input

The only line contains two integers n and k (1 ≤ n, k ≤ 109).

Output

Print the smallest integer x > n, so it is divisible by the number k.

Examples
input
5 3
output
6
input
25 13
output
26
input
26 13
output
39
题意:找到一个大于n并且被k整除的最小的数;
思路:除+1;
#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+,M=1e6+,inf=1e9+;
int main()
{
int x,y,z,i,t;
scanf("%d%d",&x,&y);
printf("%I64d\n",(ll)(x/y+)*y);
return ;
}

代码

B. The Same Calendar
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

The girl Taylor has a beautiful calendar for the year y. In the calendar all days are given with their days of week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday.

The calendar is so beautiful that she wants to know what is the next year after y when the calendar will be exactly the same. Help Taylor to find that year.

Note that leap years has 366 days. The year is leap if it is divisible by 400 or it is divisible by 4, but not by 100(https://en.wikipedia.org/wiki/Leap_year).

Input

The only line contains integer y (1000 ≤ y < 100'000) — the year of the calendar.

Output

Print the only integer y' — the next year after y when the calendar will be the same. Note that you should find the first year after y with the same calendar.

Examples
input
2016
output
2044
input
2000
output
2028
input
50501
output
50507
Note

Today is Monday, the 13th of June, 2016.

题意:找到这年之后日历相同的年份;

思路:天数整除7并且是同是闰年或平年;

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+,M=1e6+,inf=1e9+;
int check(int x)
{
if((x%==&&x%!=)||x%==)
return ;
return ;
}
int main()
{
int x,y,z,i,t;
scanf("%d",&x);
int ans=;
for(i=x;;i++)
{
ans+=check(i)%;
if(ans%==&&(check(x)==check(i+)))
break;
}
cout<<i+<<endl;
return ;
}

代码

C. Joty and Chocolate
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Joty has got a task to do. She has a line of n tiles indexed from 1 to n. She has to paint them in a strange pattern.

An unpainted tile should be painted Red if it's index is divisible by a and an unpainted tile should be painted Blue if it's index is divisible by b. So the tile with the number divisible by a and b can be either painted Red or Blue.

After her painting is done, she will get p chocolates for each tile that is painted Red and q chocolates for each tile that is painted Blue.

Note that she can paint tiles in any order she wants.

Given the required information, find the maximum number of chocolates Joty can get.

Input

The only line contains five integers nabp and q (1 ≤ n, a, b, p, q ≤ 109).

Output

Print the only integer s — the maximum number of chocolates Joty can get.

Note that the answer can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long long integer type and in Java you can use long integer type.

Examples
input
5 2 3 12 15
output
39
input
20 2 3 3 5
output
51

题意:1-n中每个整除a的数可以得到p;1-n中每个整除b的数可以得到q;

    问最多可以得到多少;

思路:简单的容斥;

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+,M=1e6+,inf=1e9+;
ll gcd(ll x,ll y)
{
return y==?x:gcd(y,x%y);
}
int main()
{
int x,y,z,i,t;
ll n,a,b,p,q;
scanf("%I64d%I64d%I64d%I64d%I64d",&n,&a,&b,&p,&q);
ll aa=n/a;
ll bb=n/b;
ll ans=aa*p+bb*q;
ll lcm=a*b/gcd(a,b);
ans-=n/lcm*min(p,q);
printf("%I64d\n",ans);
return ;
}

代码

D. Iterated Linear Function
time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Consider a linear function f(x) = Ax + B. Let's define g(0)(x) = x and g(n)(x) = f(g(n - 1)(x)) for n > 0. For the given integer values AB,n and x find the value of g(n)(x) modulo 109 + 7.

Input

The only line contains four integers ABn and x (1 ≤ A, B, x ≤ 109, 1 ≤ n ≤ 1018) — the parameters from the problem statement.

Note that the given value n can be too large, so you should use 64-bit integer type to store it. In C++ you can use the long longinteger type and in Java you can use long integer type.

Output

Print the only integer s — the value g(n)(x) modulo 109 + 7.

Examples
input
3 4 1 1
output
7
input
3 4 2 1
output
25
input
3 4 3 1
output
79

思路:赤裸裸的矩阵快速幂;

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e3+,M=1e6+,inf=1e9+;
struct is
{
ll a[][];
};
ll x,m,k,c;
is juzhenmul(is a,is b,ll hang ,ll lie)
{
int i,t,j;
is ans;
memset(ans.a,,sizeof(ans.a));
for(i=;i<=hang;i++)
for(t=;t<=lie;t++)
for(j=;j<=lie;j++)
{
ans.a[i][t]+=(a.a[i][j]*b.a[j][t]);
ans.a[i][t]%=mod;
}
return ans;
}
is quickpow(is ans,is a,ll x)
{
while(x)
{
if(x&) ans=juzhenmul(ans,a,,);
a=juzhenmul(a,a,,);
x>>=;
}
return ans;
}
int main()
{
int y,z,i,t;
ll a,b,n,x;
scanf("%I64d%I64d%I64d%I64d",&a,&b,&n,&x);
is base;
base.a[][]=a;
base.a[][]=;
base.a[][]=b;
base.a[][]=;
is ans;
memset(ans.a,,sizeof(ans.a));
ans.a[][]=;
ans.a[][]=;
ans=quickpow(ans,base,n);
printf("%I64d\n",(x*ans.a[][]+ans.a[][])%mod);
return ;
}

代码

Educational Codeforces Round 13 A、B、C、D的更多相关文章

  1. Educational Codeforces Round 88 (Rated for Div. 2) B、New Theatre Square C、Mixing Water

    题目链接:B.New Theatre Square 题意: 你要把所有"." 都变成"*",你可以有两个选择,第一种就是一次铺一个方块(1*1),第二种就是同一 ...

  2. Educational Codeforces Round 88 (Rated for Div. 2) E、Modular Stability 逆元+思维

    题目链接:E.Modular Stability 题意: 给你一个n数,一个k,在1,2,3...n里挑选k个数,使得对于任意非负整数x,对于这k个数的任何排列顺序,然后用x对这个排列一次取模,如果最 ...

  3. Educational Codeforces Round 45 (Rated for Div. 2) C、D

      C. Bracket Sequences Concatenation Problem time limit per test 2 seconds memory limit per test 256 ...

  4. Educational Codeforces Round 92 (Rated for Div. 2) B、C题解

    TAT 第一场codeforces B. Array Walk #暴力 #贪心 题目链接 题意 有\(a1, a2, ..., an\) 个格子(每个格子有各自分数),最初为1号格(初始分数为\(a1 ...

  5. Educational Codeforces Round 88 (Rated for Div. 2) D、Yet Another Yet Another Task

    题意: 给你一个含n个数a1,a2...an的数组,你要找到一个区间[l,r],使得al+a(l+1)+...+a(r-1)+ar减去max(al,a(l+1),...,a(r-1),ar)的值尽可能 ...

  6. Educational Codeforces Round 13 D:Iterated Linear Function(数论)

    http://codeforces.com/contest/678/problem/D D. Iterated Linear Function Consider a linear function f ...

  7. Educational Codeforces Round 13 D. Iterated Linear Function (矩阵快速幂)

    题目链接:http://codeforces.com/problemset/problem/678/D 简单的矩阵快速幂模版题 矩阵是这样的: #include <bits/stdc++.h&g ...

  8. Educational Codeforces Round 13 E. Another Sith Tournament 状压dp

    E. Another Sith Tournament 题目连接: http://www.codeforces.com/contest/678/problem/E Description The rul ...

  9. Educational Codeforces Round 13 D. Iterated Linear Function 水题

    D. Iterated Linear Function 题目连接: http://www.codeforces.com/contest/678/problem/D Description Consid ...

随机推荐

  1. ubuntu14下创建软件的快捷启动方式

    下载软件,使用softname/bin/softname.sh即可启动,但是很麻烦,每次都要打开terminal 为了方便,我们需要创建desktop文件指向这个启动软件的shell文件(以创建Pyc ...

  2. HTTP API 设计指南(转)

    add by zhj (2014-12-16): 今天才知道,原画HeroKu是国外一个很有名的PaaS提供商,公司很可能会将app迁移到他们那里 英文原文: HTTP API Design Guid ...

  3. 把typora改为微软雅黑+Consolas

    前言 typora是一款非常方便的书写markdown文本的编辑器.官网:https://www.typora.io/ 对于字体强迫症患者来说,不把字体改成微软雅黑+Consolas,那是相当难受.本 ...

  4. 20165324 Java实验五 网络编程与安全

    20165324 Java实验五 网络编程与安全 一.实验报告封面 课程:Java程序设计 班级:1653班 姓名:何春江 学号:20165324 指导教师:娄嘉鹏 实验日期:2018年5月28日 实 ...

  5. 使用Echarts进行可视化的数据线呈现

    由于游戏后台需要统计游戏玩家的支付情况,恰好那天看见同学群里聊天说到了Echarts,于是我就看了眼,一看,哟,还是百度的产品,看了文档,示例,确实很屌的样子啊,于是自己就开始试了,最终效果如下: 个 ...

  6. H5端js实现图片放大滑动查看-插件photoswipe的使用

    最近在开发项目的时候,遇到一个需求,需要移动端实现放大查看图片的功能,然后我就在网上搜索了一下资料,看到了photoswipe这个插件,后来试了试,确实挺好用的,它可以实现手势放大缩小查看图片,左右滑 ...

  7. [笔记] Ubuntu 18.04安装Docker CE及nvidia-docker2流程

    Docker的好处之一,就是在Container里面可以随意瞎搞,不用担心弄崩Host的环境. 而nvidia-docker2的好处是NVidia帮你配好了Host和Container之间的CUDA相 ...

  8. windows安装oracle client 18c 和plsql工具

    安装须知: (1)安装平台选择.linux/windows (2)软件位数选择.32/64,如果你的plsql工具是32位,那么你就安装32位客户端,如果是64位,你就安装64位客户端. 安装过程: ...

  9. exec-maven-plugin配置及使用

    背景: 如果你想在项maven生命周期内,运行一段java代码,或者一段独立的程序,或者说我们所指的预执行,初始化某些值,生成某些不能预先生成的文件.那么这样我们就可以使用exec-maven-plu ...

  10. linux_rpm命令

    rpm 常用命令 1.安装一个包 # rpm -ivh 文件名 2.升级一个包 # rpm -Uvh文件名 3.移走一个包 # rpm -e文件名 4.安装参数 --force 即使覆盖属于其它包的文 ...