HDU 6153 A Secret

A Secret

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)

Total Submission(s): 1924    Accepted Submission(s): 704

Problem Description

Today is the birthday of SF,so VS gives two strings S1,S2 to SF as a present,which have a big secret.SF is interested in this secret and ask VS how to get it.There are the things that VS tell:
  Suffix(S2,i) = S2[i...len].Ni is the times that Suffix(S2,i) occurs in S1 and Li is the length of Suffix(S2,i).Then the secret is the sum of the product of Ni and Li.
  Now SF wants you to help him find the secret.The answer may be very large, so the answer should mod 1000000007.

 

Input

Input contains multiple cases.
  The first line contains an integer T,the number of cases.Then following T cases.
  Each test case contains two lines.The first line contains a string S1.The second line contains a string S2.
  1<=T<=10.1<=|S1|,|S2|<=1e6.S1 and S2 only consist of lowercase ,uppercase letter.

 

Output

For each test case,output a single line containing a integer,the answer of test case.
  The answer may be very large, so the answer should mod 1e9+7.

 

Sample Input

2aaaaaaaabababababa

 

Sample Output

1319

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

题意：给定两个字符串T和P，问P的所有后缀的长度与其在字符串T中出现次数的乘积的和对1e9+7取余的结果。

解题思路：问的是后缀的匹配，其实将两个字符串都翻转一次就变成了前缀的匹配，而前缀的匹配问题就是扩展KMP的应用，所以想到这个就直接是KMP求next加扩展KMP求前缀匹配的问题了，因为问的是P的所有后缀的长度与其在字符串T中出现次数的乘积的和对1e9+7取余的结果，仔细一想其实就是T的后缀与P的最长公共前缀的数组extend的一个等差数列求和，如输入T=abababab，P=aba，翻转后T=babababa，P=aba，T的扩展数组extend={0,3,0,3,0,3,0,1}，extend[1]=3，说明T中有字串aba，这一个子串中a出现一次，ba出现一次，aba出现一次，所以在这一子串中的部分和为3+2+1，就是extend[1]=3作为末项，1为首项公差为1的一个等差数列求和，所以总的答案就是所有不为零的extend作为等差数列的末项1为首项公差为1求和。

AC代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;const int MAXL = 1e6 + 10;const int MOD = 1e9 + 7;char T[MAXL];char P[MAXL];int fail[MAXL];ll extend[MAXL];void get_fail(int m) //子串本身的最长公共前缀数组 {memset(fail,0,sizeof(fail));fail[0]=m; //与自身的最长公共前缀为本身的长度 int p,a; //p为匹配到达的最远位置，a为达到最远位置的起始位置 for(int i=1,j=-1;i<m;i++,j--) {//第一种情况，i+fail[i-a]>=p if(j<0 || i+fail[i-a]>=p) {if(j<0) j=0,p=i;while(p<m && P[j]==P[p]){j++;p++;}fail[i]=j;a=i;}//第二种情况：i+fail[i-a]<p else fail[i]=fail[i-a];}}void get_extend(int n,int m) //解释同get_fail() {memset(extend,0,sizeof(extend));int a,p;for(int i=0,j=-1;i<n;i++,j--){if(j<0 || i+fail[i-a]>=p){if(j<0) j=0,p=i;while(p<n && j<m && T[p]==P[j]){j++;p++;}extend[i]=j;a=i;}else extend[i]=fail[i-a];}}ll count(int n){ll sum=0;for(int i=0;i<n;i++){if(extend[i]!=0) //不为零的extend[i]，作为末项，1为首项公差为1求和 {ll item=(extend[i]*(extend[i]+1)/2)%MOD;sum=(sum+item)%MOD;}}return sum; //返回结果 }void solve(){int lenT=strlen(T); //母串的长度 int lenP=strlen(P); //子串的长度 reverse(T,T+lenT); //将字符串翻转 reverse(P,P+lenP); get_fail(lenP); //获取子串本身的最长公共前缀数组 get_extend(lenT,lenP); //获得母串对于子串的最长公共前缀数组 ll ans=count(lenT); //计算结果 printf("%lld\n",ans);}int main(void){int test; //测试组数 scanf("%d",&test);getchar();while(test--){gets(T); //输入母串 gets(P); //输入字串 solve();  //处理函数 } return 0;}