題目鏈接：https://codeforces.com/problemset/problem/1082/A

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題目意思：

一本書有1到n頁，起始頁數是x，

一次翻頁只能翻d（向前或者向后翻，但是不能翻出去，例如n=5,x=1? y=5,d=10? ? ?）? ?這種情況下是可以翻頁到5的。

問能否翻y頁。否輸出-1，能則輸出從x到y所需的最少翻頁次數

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思路：

1.　　x能直接到y。

2.　　起點x非1且非n，同時d>=n。

3.　　x先到1，再到y。

4.　　x先到n，再到y。　　　　

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AC代碼如下：

#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<cstdlib> #include<cmath> #include<algorithm> #include<vector> #include<queue> #include<stack> #include<set> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define Swap(a,b,t) t=a,a=b,b=t #define Mem0(x) memset(x,0,sizeof(x)) #define Mem1(x) memset(x,-1,sizeof(x)) #define MemX(x) memset(x,0x3f,sizeof(x)) using namespace std; typedef long long ll; const ll inf=0xfffffffff; const double eps=1e-12; int main() {int t;cin>>t;while(t--){ll n,x,y,d;cin>>n>>x>>y>>d;if (x==y){printf("0\n");continue;}else if (abs(x-y)%d==0){printf("%d\n",abs(x-y)/d);continue;}else if ((y!=1&&y!=n)&&d>=n){printf("-1\n");continue;}else{ //x->1->yll ans1=0,ans2=0,ans;ans1=ans1+(x-1)/d;if ((x-1)%d!=0){ans1+=1;}if ((y-1)%d!=0)ans1=inf;else ans1=ans1+(y-1)/d;//x->n->1ans2=ans2+(n-x)/d;if ((n-x)%d!=0)ans2++;if((n-y)%d!=0)ans2=inf;elseans2=ans2+(n-y)/d; // printf("%lld\n\n",ans2);ans=min(ans1,ans2);if (ans==inf)printf("-1\n");else printf("%lld\n",ans); // printf("inf=%lld",inf); }} return 0; }

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A. Vasya and Book time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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.

Input

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.

Output

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.

Example input Copy 3 10 4 5 2 5 1 3 4 20 4 19 3 output Copy 4 -1 5 Note

In the first test case the optimal sequence is:?4→2→1→3→54→2→1→3→5.

In the second test case it is possible to get to pages?11?and?55.

In the third test case the optimal sequence is:?4→7→10→13→16→194→7→10→13→16→19.

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轉載于:https://www.cnblogs.com/q1204675546/p/10060549.html

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