ZOJ Problem Set - 3818
Pretty Poem
Time Limit: 2 Seconds
Memory Limit: 65536 KB
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 symbol
A, 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
3 niconiconi~ pettan,pettan,tsurupettan wafuwafu
Sample Output
Yes Yes No
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 55
char 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;
}
/*
2
ababa
ababcab
*/