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求最小公倍数考虑得太复杂了。。。矩阵行列互不影响诶
可以先把每行看成一个元素求循环节,再把每列看成一个元素求循环节。。。然后相乘即可。
附简单易懂暴力代码(慢,因为string很慢)
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<string>
#include<iostream>
using namespace std;
int r,c,rl=1,cl=1;
string g[11000],g2[11000];
int mcs[11000];
int main()
{
scanf("%d%d",&r,&c);
for(int i=1;i<=r;i++)
cin>>g[i];
int t=1,tmcs=-1;mcs[0]=-1;
for(int i=1;i<=r;i++)
{
while(tmcs!=-1)
{
if(g[i].compare(g[tmcs+1])==0)
{
tmcs++;mcs[i]=tmcs;break;
}
else
tmcs=mcs[tmcs];
}
if(tmcs==-1)
{
mcs[i]=0;tmcs=0;
}
}
for(int i=1;i<=r;i++)
{
if(g[i].compare(g[t])==0)
{
t=t%rl+1;
}
else
{
rl=i-mcs[i];
t=mcs[i]%rl+1;
}
}
for(int i=1;i<=r;i++)
for(int j=0;j<c;j++)
g2[j+1]+=g[i][j];
memset(mcs,0,sizeof(mcs));
t=1,tmcs=-1;mcs[0]=-1;
for(int i=1;i<=c;i++)
{
while(tmcs!=-1)
{
if(g2[i].compare(g2[tmcs+1])==0)
{
tmcs++;mcs[i]=tmcs;break;
}
else
tmcs=mcs[tmcs];
}
if(tmcs==-1)
{
mcs[i]=0;tmcs=0;
}
}
for(int i=1;i<=c;i++)
{
if(g2[i].compare(g2[t])==0)
{
t=t%cl+1;
}
else
{
cl=i-mcs[i];
t=mcs[i]%cl+1;
}
}
printf("%d",rl*cl);
return 0;
}
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