ZOJ 3675 Trim the Nails 小水题

Trim the Nails
Time Limit: 2 Seconds Memory Limit: 65536 KB
Robert is clipping his fingernails. But the nail clipper is old and the edge of the nail clipper is potholed.

The nail clipper's edge is N millimeters wide. And we use N characters('.' or '*') to represent the potholed nail clipper. '.' represents 1 bad millimeter edge, and '*' represents 1 good millimeter edge.(eg. "*****" is a 5 millimeters nail clipper with the whole edge good. "***..." is a 6 millimeters nail clipper with half of its edge good and half of its edge bad.)

Notice Robert can turn over the clipper. Turning over a "**...*"-nail clipper will make a "*...**"-nail clipper.

One-millimeter good edge will cut down Robert's one-millimeter fingernail. But bad one will not. It will keep the one-millimeter unclipped.

Robert's fingernail is M millimeters wide. How many times at least should Robert cut his fingernail?

Input
There will be multiple test cases(about 15). Please process to the end of input.

First line contains one integer N.(1≤N≤10)

Second line contains N characters only consists of '.' and '*'.

Third line contains one integer M.(1≤M≤20)

Output
One line for each case containing only one integer which is the least number of cuts. If Robert cannot clipper his fingernail then output -1.

Sample Input
8
****..**
4
6
*..***
7
Sample Output
1
2
Hint
We use '-' to present the fingernail.
For sample 1:
fingernail: ----
nail clipper: ****..**
Requires one cut.

For sample 2:
fingernail: -------
nail clipper: *..***
nail clipper turned over: ***..*
Requires two cuts.

题意：一把剪刀有豁口 .表示豁口 *表示能用的地方用剪刀剪指甲指甲是平的问最少需要多少次把所有的指甲剪掉

输入n表示剪刀长度输入n个字符输入m表示指甲长度
比赛的时候只瞟了一眼也没仔细思考就直接搞别的出题率高的题目去了浪费了白花花的积分啊主要原因还是自己速度太慢
要自信要速度做题遇到题目有了想法不要畏手畏脚要敢于拍出来不行就果断放弃给别的题目腾出时间

另外做题中出现了好几个细节错误这些错误本应该在拍代码中避免的但是自己总是不够严谨细心以后要特意去加强下

还要加强代码能力越来越发现代码能力的重要性了

[cpp]
#include<stdio.h>
#include<string.h>
int a[100];

int n,m,min,left,right;
void DFS(int b[],int s,int flag,int step)
{
    int i,j,c[100];
    if(step>=min) return;
    if(flag)
    {
        for(i=1;i<=25;i++)
            c[i]=b[i];
        for(i=s,j=left;i<=m&&j<=n;i++,j++)
            if(c[i]==0)
            c[i]=a[j];
        for(i=s;i<=m;i++)
            if(c[i]==0)
            {
                s=i;break;
            }
            if(i==m+1)
            {
                if(step<min) min=step;
                return ;
            }
            DFS(c,s,1,step+1);
            DFS(c,s,0,step+1);
    }
    else
    {
        for(i=1;i<=25;i++)
            c[i]=b[i];
        for(i=s,j=right;i<=m&&j>0;i++,j--)
            if(c[i]==0)
            c[i]=a[j];
        for(i=s;i<=m;i++)
            if(c[i]==0)
            {
                s=i;break;
            }
            if(i==m+1)
            {
                if(step<min) min=step;
                return ;
            }
            DFS(c,s,0,step+1);
            DFS(c,s,1,step+1);
    }
}
int main()
{
    int i,b[100];
    char ch;
    while(scanf("%d",&n)!=EOF)
    {
        getchar();//这里要有getchar 否则会影响下面的字符输入
        for(i=1;i<=n;i++)
        {
            scanf("%c",&ch);
            if(ch=='*') a[i]=1;
            else a[i]=0;
        }
        left=0;
        for(i=1;i<=n;i++)
            if(a[i]==1)
            {
                left=i;
                break;
            }
            for(i=n;i>0;i--)
          &

补充：软件开发 , C++ ,