Vasya and Book

题目链接

Vasya is reading a e-book. The file of the book consists of n pages, numbered from 1 to n. The screen is currently displaying the contents of page x, and Vasya wants to read the page y. There are two buttons on the book which allow Vasya to scroll d pages forwards or backwards (but he cannot scroll outside the book). For example, if the book consists of 10 pages, and d=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 y.

Input
The first line contains one integer t (1≤t≤10³) — the number of testcases.

Each testcase is denoted by a line containing four integers n, x, y, d (1≤n,d≤10⁹, 1≤x,y≤n) — the number of pages, the starting page, the desired page, and the number of pages scrolled by pressing one button, respectively.

Output
Print one line for each test.

If Vasya can move from page x to page y, print the minimum number of times he needs to press a button to do it. Otherwise print −1.

input

3
10 4 5 2
5 1 3 4
20 4 19 3

output

4
-1
5

Note
In the first test case the optimal sequence is: 4→2→1→3→5.

In the second test case it is possible to get to pages 1 and 5.

In the third test case the optimal sequence is: 4→7→10→13→16→19.

 

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<cmath>
 6 using namespace std;
 7 int main(){
 8     int t,n,x,y,d,tmp,min_page;
 9     scanf("%d",&t);
10     while (t--){
11         tmp = 0;
12         min_page = 1e+9;
13         scanf("%d%d%d%d",&n,&x,&y,&d);
14         //第一种情况:可直接翻过
15         if (abs(x - y) % d == 0)
16             min_page = min(min_page,abs(x - y) / d);
17         //第二种情况:先往前翻页，再往后翻页
18         if ((y - 1) % d == 0){
19             tmp += (x - 1) / d + (y - 1) / d;
20             if((x - 1) * 1.0 / d != 0)
21                 tmp++;
22             min_page = min(min_page,tmp);
23         }
24         //第三种情况:先往后翻页，再往前翻页
25         tmp = 0;
26         if ((n - y) % d == 0){
27             tmp += (n - y) / d + (n - x) / d;
28             if((n - x) * 1.0 / d != 0)
29                 tmp++;
30             min_page = min(min_page,tmp);
31         }
32         if (min_page == 1e+9)
33             printf("-1\n");
34         else
35             printf("%d\n",min_page);
36     }
37     return 0;
38 }

View Code

 

 

转载于:https://www.cnblogs.com/Luckykid/p/10054948.html