hdoj 5523 Game 【细心题】

Game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 458    Accepted Submission(s): 182

Problem Description

XY is playing a game:there are N pillar in a row,which numbered from 1 to n.Each pillar has a jewel.Now XY is standing on the S-th pillar and the exit is in the T-th pillar.XY can leave from the exit only after they get all the jewels.Each time XY can move to adjacent pillar,or he can jump to boundary ( the first pillar or the N-th pillar) by using his superpower.However,he needs to follow a rule:if he left the pillar,he no can not get here anymore.In order to save his power,XY wants to use the minimum number of superpower to pass the game.

 

Input

There are multiple test cases, no more than 1000 cases.
For each case,the line contains three integers:N,S and T.(1≤N≤10000,1≤S,T≤N)

 

Output

The output of each case will be a single integer on a line: the minimum number of using superpower or output -1 if he can't leave.

 

Sample Input

4 1 44 1 3

 

Sample Output

01

 

题意：给定n根排成一行的柱子，每个柱子上都有一个宝石。你拿走所有宝石才可以从出口离开，且每根柱子只能走一次。你可以选择用超能力跳的编号为1或者编号为n的柱子，现在给你入口和出口，问你离开需要使用超能力的最少次数，若不能离开输出-1。

思路：细心点就好了，注意1 1 1 结果是 0。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#define MAXN 110using namespace std;int main(){    int N, S, T;    while(scanf("%d%d%d", &N, &S, &T) != EOF)    {        if(S == T)        {            if(N == 1)                printf("0\n");            else                printf("-1\n");        }        else if(S == 1 && T == N || S == N && T == 1)            printf("0\n");        else if(S == 1 || S == N)            printf("1\n");        else if(abs(S-T) == 1)            printf("1\n");        else            printf("2\n");    }    return 0;}