Description：
遇到了ogo可以变成***如果ogo后面有go统统忽略，输出结果
Solution:
哎如果我一开始对题意的解读如上的话，就不会被整的那么麻烦了
Code：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 150;
/***找到一个题目的规律，做题的方法是多么的重要！！***/
/***以后没有好的方法，不打代码***/
char s[maxn];
int main()
{
int len;
while(~scanf("%d%s",&len,s))
{
for(int i = 0;i < len;)
{
if(s[i] == 'o' && s[i + 1] == 'g' && s[i + 2] == 'o')
{
printf("***");
i += 3;
while(s[i] == 'g' && s[i+1] == 'o')
{
i += 2;
}
}
else
{
printf("%c",s[i]);
i++;
}
}
printf("\n");
}
return 0;
}
看看我的超级麻烦bug多多的模拟！！！
一开始真的读错题了！
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm> using namespace std;
const int maxn = 1e3;
char s[maxn];
char out[maxn];
char cmp[200] = "ogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogogo";
int main()
{
int len;
while(~scanf("%d",&len))
{
scanf("%s",s);
int mpid = 0;
int outsid = 0;
for(int i = 0;i < len;++i)
{
out[outsid++] = s[i];
if(s[i] == cmp[mpid])
{
//printf(" %d\n",mpid);
mpid++;
}
else
{
//cout<<mpid<<endl;
if(mpid == 1)
{
outsid -= 1;
i -= 1;
}
else if(mpid == 2)
{
mpid = 0;
}
else if(mpid % 2 == 1)
{
outsid -= mpid + 1;
out[outsid++] = '*';
out[outsid++] = '*';
out[outsid++] = '*';
i--;
}
else if(mpid % 2 == 0 && mpid != 0)
{
outsid -= mpid + 1;
out[outsid++] = '*';
out[outsid++] = '*';
out[outsid++] = '*';
out[outsid++] = 'g';
out[outsid++] = s[i];
}
mpid = 0;
}
}
if(mpid > 2)
{
if(mpid % 2 == 1)
{
outsid -= mpid;
out[outsid++] = '*';
out[outsid++] = '*';
out[outsid++] = '*';
}
else if(mpid % 2 == 0)
{
outsid -= mpid;
out[outsid++] = '*';
out[outsid++] = '*';
out[outsid++] = '*';
out[outsid++] = 'g';
}
}
out[outsid] = '\0';
printf("%s\n",out);
}
return 0;
}