<Codeforce>1082A. Vasya and Book
Vasya is reading a e-book. The file of the book consists of nn pages, numbered from 11 to nn. The screen is currently displaying the contents of page xx, and Vasya wants to read the page yy. There are two buttons on the book which allow Vasya to scroll dd pages forwards or backwards (but he cannot scroll outside the book). For example, if the book consists of 1010 pages, and d=3d=3, then from the first page Vasya can scroll to the first or to the fourth page by pressing one of the buttons; from the second page — to the first or to the fifth; from the sixth page — to the third or to the ninth; from the eighth — to the fifth or to the tenth.
Help Vasya to calculate the minimum number of times he needs to press a button to move to page yy.
The first line contains one integer tt (1≤t≤1031≤t≤103) — the number of testcases.
Each testcase is denoted by a line containing four integers nn, xx, yy, dd (1≤n,d≤1091≤n,d≤109, 1≤x,y≤n1≤x,y≤n) — the number of pages, the starting page, the desired page, and the number of pages scrolled by pressing one button, respectively.
Print one line for each test.
If Vasya can move from page xx to page yy, print the minimum number of times he needs to press a button to do it. Otherwise print −1−1.
INPUT:
3
10 4 5 2
5 1 3 4
20 4 19 3
OUTPUT:
4
-1
5
记得fmax和fmin要开G++编译器!
#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
int main(){
int t,p,a,b,x;
cin>>t;
while(t--){
long long sum=0;
cin>>p>>a>>b>>x;
if(a==b){
cout<<0<<endl;
}else{
int k=abs(a-b); if(k%x==0){
sum=k/x;
cout<<sum<<endl;
}else{
/*两种情况,a->1->b,a->p->b*/
int m1,m2;
if((a-1)%x==0){
sum+=(a-1)/x;
}else{
sum+=(a-1)/x+1;
}
if((b-1)%x==0){
sum+=(b-1)/x;
}else{
sum=-1;
}
m1=sum;
sum=0;
if((p-a)%x==0){
sum+=(p-a)/x;
}else{
sum+=(p-a)/x+1;
}
if((p-b)%x==0){
sum+=(p-b)/x;
}else{
sum=-1;
}
m2=sum;
//cout<<m1<<endl<<m2<<endl<<endl;
if(m1!=-1&&m2!=-1){
if(m1<m2){
cout<<m1<<endl;
}else{
cout<<m2<<endl;
}
//cout<<fmin(m1,m2)<<endl;
}if(m1!=-1&&m2==-1){
cout<<m1<<endl;
}if(m1==-1&&m2!=-1){
cout<<m2<<endl;
}if(m1==-1&&m2==-1){
cout<<-1<<endl;
}
}
}
}
return 0;
}
<Codeforce>1082A. Vasya and Book的更多相关文章
- CodeForce Educational round Div2 C - Vasya and Robot
http://codeforces.com/contest/1073/problem/C 题意:给你长度为n的字符串,每个字符为L, R, U, D.给你终点位置(x, y).你每次出发的起点为( ...
- Milliard Vasya's Function-Ural1353动态规划
Time limit: 1.0 second Memory limit: 64 MB Vasya is the beginning mathematician. He decided to make ...
- Codeforce 493c
H - Vasya and Basketball Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & ...
- CF460 A. Vasya and Socks
A. Vasya and Socks time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforce - Street Lamps
Bahosain is walking in a street of N blocks. Each block is either empty or has one lamp. If there is ...
- 递推DP URAL 1353 Milliard Vasya's Function
题目传送门 /* 题意:1~1e9的数字里,各个位数数字相加和为s的个数 递推DP:dp[i][j] 表示i位数字,当前数字和为j的个数 状态转移方程:dp[i][j] += dp[i-1][j-k] ...
- Codeforce Round #216 Div2
e,还是写一下这次的codeforce吧...庆祝这个月的开始,看自己有能,b到什么样! cf的第二题,脑抽的交了错两次后过了pretest然后system的挂了..脑子里还有自己要挂的感觉,果然回头 ...
- Codeforces Round #281 (Div. 2) D. Vasya and Chess 水
D. Vasya and Chess time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #281 (Div. 2) C. Vasya and Basketball 二分
C. Vasya and Basketball time limit per test 2 seconds memory limit per test 256 megabytes input stan ...
随机推荐
- hdu4313 贪心+并查集
题意简单思路也还可以.开始从小到大排序非常烦.后来从大到小就很简单了: 从大到小解决了删除的边最小. #include<stdio.h> #include<string.h> ...
- [Linux]环境配置之jdk的安装 标签: jdk服务器linux 2016-08-07 22:18 502人阅读 评论(21)
这两天服务器崩了,所以需要重新配置环境,然后从头到尾配置了一遍,现在记录总结一下自己这两天的工作,首先是jdk的配置! 很多软件,需要jdk为基础,所以第一个装的就是jdk. 第一步,拷贝文件 首先将 ...
- 【NS2】学习点滴
1 $ns duplex-link-op $n2 $n3 queuePos 0.5#此命令用于设置在NAM中显示的队列方向#经测试,发现: # queuePos 0.5表示包从上到下进入队列# que ...
- 在JS中模拟表单的post提交,进行页面的跳转
原文链接:https://blog.csdn.net/jal517486222/article/details/83147761 /* *功能: 模拟form表单的提交 *参数: URL 跳转地址 P ...
- LightOJ 1236 Pairs Forming LCM【整数分解】
题目链接: http://lightoj.com/login_main.php?url=volume_showproblem.php?problem=1236 题意: 找与n公倍数为n的个数. 分析: ...
- 随机数专题 Day08
package com.sxt.arraytest2; import java.util.Arrays; /* * 随机数专题 * Math类的random()方法 * m~n的随机数 * 公式:(i ...
- Java练习 SDUT-1140_面向对象程序设计上机练习一(函数重载)
面向对象程序设计上机练习一(函数重载) Time Limit: 1000 ms Memory Limit: 65536 KiB Problem Description 利用数组和函数重载求5个数最大值 ...
- PHP header 的7种用法
这篇文章介绍的内容是关于PHP header()的7种用法 ,有着一定的参考价值,现在分享给大家,有需要的朋友可以参考一下 PHP header 的7种用法 1. 跳转页面 header('Locat ...
- deepin golang微服务搭建go-micro环境
1.安装micro 需要使用GO1.11以上版本 #linux 下 export GO111MODULE=on export GOPROXY=https://goproxy.cn # 使用如下指令安装 ...
- Javascript 严格模式下不允许删除一个不允许删除的属性
如下代码,在严格模式下,如果删除 Object.prototype 浏览器会报错,目前 IE10 也支持 严格模式. <script> "use strict"; de ...