[NOI2014]动物园
题解:
先求出可重复的nex[]和num[].然后再次做kmp时,对于一个位置i一直跳到失配位置j,且(j<<1)<=i,所以此时重叠的num[i]=此时的num[j]+1.
1 #include<iostream> 2 #include<cstdlib> 3 #include<cstdio> 4 #include<cmath> 5 #include<algorithm> 6 #include<cstring> 7 #include<queue> 8 #include<vector> 9 #include<stack> 10 #define MAXN 1000010 11 #define RG register 12 #define LL long long int 13 using namespace std; 14 const int INF=1e9; 15 const int mod=1e9+7; 16 int T; 17 int n; 18 char s[MAXN]; 19 int nex[MAXN],num[MAXN]; 20 LL ans; 21 void work() 22 { 23 int j=0;ans=1;num[1]=1; 24 for(int i=2;i<=n;i++) 25 { 26 while(j>0&&s[j+1]!=s[i]) j=nex[j]; 27 if(s[j+1]==s[i]) j++; 28 nex[i]=j;num[i]=num[j]+1; 29 } 30 j=0; 31 for(int i=2;i<=n;i++) 32 { 33 while(j>0&&s[j+1]!=s[i]) j=nex[j]; 34 if(s[j+1]==s[i]) j++; 35 while((j<<1)>i&&nex[j]) j=nex[j]; 36 ans=ans*(num[j]+1)%mod; 37 } 38 } 39 int main() 40 { 41 freopen("1.in","r",stdin); 42 scanf("%d",&T); 43 while(T--) 44 { 45 scanf("%s",s+1);n=strlen(s+1); 46 work(); 47 printf("%lld\n",ans); 48 } 49 return 0; 50 }
