ACM 第十一天
多校7题目
GuGuFishtion
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1008 Accepted Submission(s): 175
At the break time, an evil idea arises in XianYu's mind.
‘Come on, you xxxxxxx little guy.’
‘I will give you a function ϕ(x) which counts the positive integers up to x that are relatively prime to x.’
‘And
now I give you a fishtion, which named GuGu Fishtion, in memory of a
great guy named XianYu and a disturbing and pitiful guy GuGu who will be
cooked without solving my problem in 5 hours.’
‘The given fishtion is defined as follow:
And now you, the xxxxxxx little guy, have to solve the problem below given m,n,p.’
So SMART and KINDHEARTED you are, so could you please help GuGu to solve this problem?
‘GU GU!’ GuGu thanks.
1≤T≤3
1≤m,n≤1,000,000
max(m,n)<p≤1,000,000,007
And given p is a prime.
5 7 23
Sequence
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1782 Accepted Submission(s): 377
Your job is simple, for each task, you should output Fn module 109+7.
Then, for the next T lines, each line consists of 6 integers, A , B, C, D, P, n.
1≤T≤200≤A,B,C,D≤1091≤P,n≤109
3 3 2 1 3 5
3 2 2 2 1 4
24
A - Round House
Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n. Entrance n and entrance 1 are adjacent.
Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance a and he decided that during his walk he will move around the house b entrances in the direction of increasing numbers (in this order entrance n should be followed by entrance 1). The negative value of b corresponds to moving |b| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance n). If b = 0, then Vasya prefers to walk beside his entrance.
Illustration for n = 6, a = 2, b = - 5.
Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.
The single line of the input contains three space-separated integers n, a and b (1 ≤ n ≤ 100, 1 ≤ a ≤ n, - 100 ≤ b ≤ 100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.
Output
Print a single integer k (1 ≤ k ≤ n) — the number of the entrance where Vasya will be at the end of his walk.
Examples
6 2 -5
3
5 1 3
4
3 2 7
3
Note
The first example is illustrated by the picture in the statements.
#include<stdio.h> int main()
{
int n;
int a,b;
int ans;
scanf("%d%d %d",&n,&a,&b);
if(b<)
{
int s=b%n;
if(s==) ans=a;
else{
ans=a+(n+(b%n));
if(ans>n)
ans=ans%n;}
}
else if(b==) ans=a;
else
{ ans=a+(b%n);
if(ans>n) ans=ans%n;
}
printf("%d\n",ans);
return ;
}
ACM 第十一天的更多相关文章
- 丁酉年六月十一ACM模拟赛
似乎该写题解了.今天模拟ACM,10道题(本来还有2道被删了),9道都来自BZOJ,中间我做过2道.那么说,今天Solv.便大大增多了(但还是不如强大的Amphetamine). 题单及一句话题解如下 ...
- 【ACM】魔方十一题
0. 前言打了两年的百度之星,都没进决赛.我最大的感受就是还是太弱,总结起来就是:人弱就要多做题,人傻就要多做题.题目还是按照分类做可能效果比较好,因此,就有了做几个系列的计划.这是系列中的第一个,解 ...
- kuangbin专题专题十一 网络流 POJ 3436 ACM Computer Factory
题目链接:https://vjudge.net/problem/POJ-3436 Sample input 1 3 4 15 0 0 0 0 1 0 10 0 0 0 0 1 1 30 0 1 2 1 ...
- ACM题集以及各种总结大全!
ACM题集以及各种总结大全! 虽然退役了,但是整理一下,供小弟小妹们以后切题方便一些,但由于近来考试太多,顾退役总结延迟一段时间再写!先写一下各种分类和题集,欢迎各位大牛路过指正. 一.ACM入门 关 ...
- 记第五届山东省ACM程序设计比赛——遗憾并非遗憾
记第五届山东省ACM程序设计比赛 5月10日上午9点半左右,我们的队伍从学校出发,一个多小时后到达本次比赛的地点-哈尔滨工业大学. 报道,领材料,吃午饭,在哈工大的校园里逛了逛,去主楼的自习室歇息了一 ...
- ACM题集以及各种总结大全(转)
ACM题集以及各种总结大全! 虽然退役了,但是整理一下,供小弟小妹们以后切题方便一些,但由于近来考试太多,顾退役总结延迟一段时间再写!先写一下各种分类和题集,欢迎各位大牛路过指正. 一.ACM入门 关 ...
- ACM入门步骤(一)
一般的入门顺序: 0. C语言的基本语法(或者直接开C++也行,当一个java选手可能会更受欢迎,并且以后工作好找,但是难度有点大),[参考书籍:刘汝佳的<算法竞赛入门经典>,C++入门可 ...
- 牛人的ACM经验 (转)
一:知识点 数据结构: 1,单,双链表及循环链表 2,树的表示与存储,二叉树(概念,遍历)二叉树的 应用(二叉排序树,判定树,博弈 ...
- ACM算法锦集
一:知识点 数据结构: 1,单,双链表及循环链表 2,树的表示与存储,二叉树(概念,遍历)二叉树的 应用(二叉排序树,判定树,博弈树,解答树等) 3,文件操作(从文本文件中读入数据并输出到文本文 件中 ...
随机推荐
- 编写radware的负载配置
radware如何添加负载服务? 笔者在新添加radware的新负载服务的时候,是习惯去看下上一个负载服务的ID 和 节点服务的ID 号 分别是多少,主要是避免ID冲突,把其他服务顶替下去,同时以后这 ...
- 慕课笔记-JavaScript正则表达式
目录 慕课笔记-JavaScript正则表达式笔记 概述 RegExp对象 修饰符 元字符 字符类 范围类 预定义类 预定义字符 边界 量词 贪婪模式 分组 或(使用竖线表示) 反向引用 忽略分组 前 ...
- PHP Mysql数据库连接
1,date_default_timezone_set('PRC');//获取北京时区 header("Content-Type:text/html;charset=utf-8&q ...
- java中子类会继承父类的构造方法吗?
参考: https://blog.csdn.net/wangyl_gain/article/details/49366505
- Qt上FFTW組件的编译与安裝
Qt上FFTW組件的編譯安裝 FFTW是一個做頻譜非常實用的組件,本文講述在Windows和Linux兩個平臺使用FFTW組件.Windows下的的FFTW組件已經編譯好成爲dll文件,按照開發應用的 ...
- python 爬虫(爬取网页的img并下载)
from urllib.request import urlopen # 引用第三方库 import requests #引用requests/用于访问网站(没安装需要安装) from pyquery ...
- ACM1003:Max Sum
Problem Description Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum ...
- C++编译错误杂记
目录 2018年12月23日 error: no matching function for call to ××× 2018年12月10日 error: expected ')' before '* ...
- LeetCode: 29. Divide Two Integers (Medium)
1. 原题链接 https://leetcode.com/problems/divide-two-integers/description/ 2. 题目要求 给出被除数dividend和除数divis ...
- 使用CRF做命名实体识别(三)
摘要 本文主要是对近期做的命名实体识别做一个总结,会给出构造一个特征的大概思路,以及对比所有构造的特征对结构的影响.先给出我最近做出来的特征对比: 目录 整体操作流程 特征的构造思路 用CRF++训练 ...