J题 游戏模拟

J：Ostap and Grasshopper

//左右走次虫子
On the way to Rio de Janeiro Ostap kills time playing with a grasshopper he took with him in a special box. Ostap builds a line of length n such that some cells of this line are empty and some contain obstacles. Then, he places his grasshopper to one of the empty cells and a small insect in another empty cell. The grasshopper wants to eat the insect.

Ostap knows that grasshopper is able to jump to any empty cell that is exactly k cells away from the current (to the left or to the right). Note that it doesn’t matter whether intermediate cells are empty or not as the grasshopper makes a jump over them. For example, if k = 1 the grasshopper can jump to a neighboring cell only, and if k = 2 the grasshopper can jump over a single cell.

Your goal is to determine whether there is a sequence of jumps such that grasshopper will get from his initial position to the cell with an insect.

Input
The first line of the input contains two integers n and k (2 ≤ n ≤ 100, 1 ≤ k ≤ n - 1) — the number of cells in the line and the length of one grasshopper’s jump.

The second line contains a string of length n consisting of characters ‘.’, ‘#’, ‘G’ and ‘T’. Character ‘.’ means that the corresponding cell is empty, character ‘#’ means that the corresponding cell contains an obstacle and grasshopper can’t jump there. Character ‘G’ means that the grasshopper starts at this position and, finally, ‘T’ means that the target insect is located at this cell. It’s guaranteed that characters ‘G’ and ‘T’ appear in this line exactly once.

Output
If there exists a sequence of jumps (each jump of length k), such that the grasshopper can get from his initial position to the cell with the insect, print “YES” (without quotes) in the only line of the input. Otherwise, print “NO” (without quotes).

Example

Input
5 2
G#T
Output
YES

Input
6 1
T….G
Output
YES

Input
7 3
T..#..G
Output
NO

Input
6 2
..GT..
Output
NO
／／／／／／／／／／／／／／／／／／／／／／／／／／／／／／／／／
Note
In the first sample, the grasshopper can make one jump to the right in order to get from cell 2 to cell 4.

In the second sample, the grasshopper is only able to jump to neighboring cells but the way to the insect is free — he can get there by jumping left 5 times.

In the third sample, the grasshopper can’t make a single jump.

In the fourth sample, the grasshopper can only jump to the cells with odd indices, thus he won’t be able to reach the insect.

note:

很简单的模拟题，大体看看题目和输入输出就大体知道题义了。
自己用while(1)实现的循环 不是题解上用的for 然后只过了5个样例。。。。
第6个奇怪的超时了应该是死循环了退不出来。

#include <iostream>#include <string>using namespace std;int main(){    string line;    int n,t,lg,lt;    bool cget=false;//判断是否能到达终点    cin>>n>>t;    for(int i=0;i<n;i++){        cin>>line[i];        if(line[i]=='G') lg=i;        if(line[i]=='T') lt=i;    }    if(lg<lt){        int pos=lg;        while(1){            pos+=t;            if(line[pos]=='T'){cget=true; break; }            if(pos>lt||line[pos]=='#') break;        }    }    else{        int pos=lg;        while(1){            pos-=t;            if(line[pos]=='T'){cget=true; break; }            if(pos<lt||line[pos]=='#') break;        }    }    if(cget) cout<<"YES"; else cout<<"NO";    return 0;}

题解代码 已AC：

#include <iostream>#include<cstdio>#include<cstring>using namespace std;int n,k;char str[110];int main(){    scanf("%d%d",&n,&k);    scanf("%s",str);    int s,e;    for(int i=0;i<n;i++){        if(str[i]=='G')            s=i;        if(str[i]=='T')            e=i;    }//输入模块    int f=0;//标记是否到达终点    for(int i=s+k;i<n;i+=k){//如果g在t的右边        if(str[i]=='#') break;        else{            if(str[i]=='T'){                f=1;                break;            }        }    }    if(f) printf("YES\n");    else{//如果g在t的左边        for(int i=s-k;i>=0;i-=k)            if(str[i]=='#')   break;            else if(str[i]=='T'){                f=1;                break;            }        if(f) printf("YES\n");        else printf("NO\n");    }    return 0;}