zoj 3818 Pretty Poem （模拟）

ZOJ Problem Set - 3818

Pretty Poem

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Poetry is a form of literature that uses aesthetic and rhythmic qualities of language. There are many famous poets in the contemporary era. It is said that a few ACM-ICPC contestants can even write poetic code. Some poems has a strict rhyme scheme like "ABABA" or "ABABCAB". For example, "niconiconi" is composed of a rhyme scheme "ABABA" with A = "ni" and B = "co".

More technically, we call a poem pretty if it can be decomposed into one of the following rhyme scheme: "ABABA" or "ABABCAB". The symbolA, B and C are different continuous non-empty substrings of the poem. By the way, punctuation characters should be ignored when considering the rhyme scheme.

You are given a line of poem, please determine whether it is pretty or not.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

There is a line of poem S (1 <= length(S) <= 50). S will only contains alphabet characters or punctuation characters.

Output

For each test case, output "Yes" if the poem is pretty, or "No" if not.

Sample Input

3niconiconi~pettan,pettan,tsurupettanwafuwafu

Sample Output

YesYesNo

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给你一个窜，取出里面的字母，问是否满足  ABABA  或  ABABCAB，形似

思路：

      分别两个函数判断，判断两种形式，ABABA  的枚举AB的长度，可以变成XYZ  形式，判断X==Y，然后从X中取出Z长度的字符串，判断是否和Z相等，还有Z与X剩余的窜必须不相等（A！=B）；

 对于ABABCAB 基本相似；

上代码了：

 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define N 55char a[N],b[N];int k;int vis[N];int judge(){    int i,j;    char x[N],y[N],z[N];    int xx,yy,zz;    for(i=2;i*2<k;i++)    {         for(j=0;j<i;j++)            {                x[j]=b[j];            }          x[j]='\0';          yy=0;          for(j;j<i*2;j++)          {              y[yy++]=b[j];          }          y[yy]='\0';          if(strcmp(x,y)!=0)  continue;          zz=0;          for(j;b[j];j++)            z[zz++]=b[j];          z[zz]='\0';          if(zz>=i) continue;          char A[N],B[N];          int AA,BB;          for(j=0;j<zz;j++)            A[j]=b[j];          A[j]='\0';          if(strcmp(A,z)!=0) continue;          BB=0;          for(j;j<i;j++)            B[BB++]=b[j];          B[BB]='\0';          if(strcmp(A,B)==0)  continue;          //printf("x=%s\ny=%s\nz=%s\n",x,y,z);          //printf("A=%s\nB=%s\n",A,B);          return 1;    }    return 0;}int judgee(){    int i,j;    int vis[N];    char x[N],y[N],z[N],w[N];    int xx,yy,zz,ww;    for(i=2;i*3<k;i++)    {        memset(vis,0,sizeof(vis));        int xx=0;        for(j=0;j<i;j++)        {            x[j]=b[j];            vis[j]=1;        }        x[j]='\0';        yy=0;        for(j;j<i*2;j++)        {            y[yy++]=b[j];            vis[j]=1;        }        y[yy]='\0';        zz=0;        j=k-i;        for(;b[j];zz++,j++)        {            vis[j]=1;            z[zz]=b[j];        }        z[zz]='\0';        ww=0;        for(j=i*2;!vis[j];j++)        {            w[ww++]=b[j];        }        w[ww]='\0';        if(strcmp(x,y)!=0) continue;        if(strcmp(y,z)!=0) continue;        char A[N],B[N];        int AA,BB;        for(j=1;j<i;j++)        {            int jj;            for(jj=0;jj<j;jj++)                A[jj]=b[jj];              A[jj]='\0';            BB=0;            for(jj;jj<i;jj++)                B[BB++]=b[jj];            B[BB]='\0';            if(strcmp(B,A)==0) continue;            if(strcmp(B,w)==0)  continue;            if(strcmp(A,w)==0)  continue;             //printf("x=%s\ny=%s\nw=%s\nz=%s\n",x,y,w,z);             //printf("A=%s\nB=%s\n",A,B);            return 1;        }    }    return 0;}int main(){   int i,t;   scanf("%d",&t);   while(t--)   {       scanf("%s",a);       int len=strlen(a);       k=0;       for(i=0;i<len;i++)        if(a[i]>='a'&&a[i]<='z'||a[i]>='A'&&a[i]<='Z')        b[k++]=a[i];       b[k]='\0';       if(judge())       {           printf("Yes\n");           continue;       }       if(judgee())       {           printf("Yes\n");           continue;       }       printf("No\n");   }   return 0;}/*2ababaababcab*/