hdu6153 A Secret CCPC1004 扩展KMP

http://acm.hdu.edu.cn/showproblem.php?pid=6153

题意：给定字符串S1和S2，求S2所有后缀在S1中出现次数与长度乘积之和。

题解：把两个字符串都翻转，处理后缀就变成了处理前缀。处理前缀就可以用到扩展KMP。处理出的extand[i]是S1[i,i+1...,len-1]与S2的最长公共前缀，那么给位置对答案的贡献就是1+2+...+extand[i]。即extand[i]*(extand[i]+1)/2。

代码：

#include<set>#include<map>#include<stack>#include<queue>#include<vector>#include<string>#include<bitset>#include<algorithm>#include<cstring>#include<cstdio>#include<cmath>#include<iomanip>#include<iostream>#define debug cout<<"aaa"<<endl#define d(a) cout<<a<<endl#define mem(a,b) memset(a,b,sizeof(a))#define LL long long#define lson l,mid,root<<1#define rson mid+1,r,root<<1|1#define MIN_INT (-2147483647-1)#define MAX_INT 2147483647#define MAX_LL 9223372036854775807i64#define MIN_LL (-9223372036854775807i64-1)using namespace std;const int N = 1000000 + 5;const LL mod = 1000000000 + 7;const double eps = 1e-8;int nxt[N],extand[N];char a[N],b[N];void Getnxt(char *T,int *nxt){int len=strlen(T),a=0;nxt[0]=len;while(a<len-1&&T[a]==T[a+1]) a++;nxt[1]=a;a=1;for(int k=2;k<len;k++){int p=a+nxt[a]-1,L=nxt[k-a];if((k-1)+L>=p){int j=(p-k+1)>0?(p-k+1):0;while(k+j<len&&T[k+j]==T[j]) j++;nxt[k]=j;a=k;}else{nxt[k]=L;}}}void GetExtand(char *S,char *T,int *nxt){Getnxt(T,nxt),mem(extand,0);int slen=strlen(S),tlen=strlen(T),a=0;int MinLen=min(slen,tlen);while(a<MinLen&&S[a]==T[a]) a++;extand[0]=a;a=0;for(int k=1;k<slen;k++){int p=a+extand[a]-1,L=nxt[k-a];if((k-1)+L>=p){int j=(p-k+1)>0?(p-k+1):0;while(k+j<slen&&j<tlen&&S[k+j]==T[j]) j++;extand[k]=j;a=k;}else{extand[k]=L;}}}int main(){int t;LL ans,temp;scanf("%d",&t);while(t--){ans=0;scanf("%s%s",a,b);reverse(a,a+strlen(a));reverse(b,b+strlen(b));GetExtand(a,b,nxt);for(int i=0;i<strlen(a);i++){if(extand[i]){temp=extand[i]%mod;temp=(((temp*(temp+1)/2)%mod)+mod)%mod;ans=(ans+temp)%mod;}}ans=(ans+mod)%mod;printf("%lld\n",ans);}return 0;}