HDU6513/CCPC2017--A Secret（KMP）

Posted 2020-10-03 Pic

A Secret

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 461    Accepted Submission(s): 182

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

2
aaaaa
aa
abababab
aba

Sample Output

13
19
Hint
case 2: 
Suffix(S2,1) = "aba",
Suffix(S2,2) = "ba",
Suffix(S2,3) = "a".
N1 = 3,
N2 = 3,
N3 = 4.
L1 = 3,
L2 = 2,
L3 = 1.
ans = （3*3+3*2+4*1）%1000000007.
 

Source

2017中国大学生程序设计竞赛 - 网络选拔赛

题意：

给出两个字符串S1,S2，求S2的所有后缀在S1中匹配到的次数与其后缀长度的乘积之和

思路：

由于乘上了后缀长度，所以对某个后缀的匹配而言，每个字母对答案的贡献是1

由此，我们可以将字符串反转，然后跑一遍kmp，在kmp的过程中统计有多少匹配，并加上他们的贡献即可

代码：

  1 /*
  2 * @FileName: D:\代码与算法\2017训练比赛\CCPC网络赛\1004-bt.cpp
  3 * @Author: Pic
  4 * @Date:   2017-08-19 16:30:09
  5 * @Last Modified time: 2017-08-19 20:54:48
  6 */
  7 #include <bits/stdc++.h>
  8 using namespace std;
  9 /*
 10 * next[]的含义： x[i-next[i]...i-1]=x[0...next[i]-1]
 11 * next[i]为满足x[i-z...i-1]=x[0...z-1]的最大z值（就是x的自身匹配）
 12 */
 13 long long sum=0;
 14 const long long mod=1e9+7;
 15 const long long MAXN=2e6+30;
 16 char x[MAXN],y[MAXN];
 17 int nxt[MAXN];
 18 void kmp_pre(char x[],int m)
 19 {
 20     long long i, j;
 21     j = nxt[0] = -1;
 22     i = 0;
 23     while (i < m)
 24     {
 25         while (-1 != j && x[i] != x[j])j = nxt[j];
 26         nxt[++i] = ++j;
 27     }
 28 }
 29 /*
 30 * kmpNext[]的意思： next‘[i]=next[next[...[next[i]]]] (直到next‘[i]<0或者
 31 x[next‘[i]]!=x[i])
 32 * 这样的预处理可以快一些
 33 */
 34 void KMP_Count( char x[],int m,  char y[],int n)
 35 {    //x是模式串， y是主串
 36     long long i, j;
 37     kmp_pre(x,m);
 38     i = j = 0;
 39     while (i < n)
 40     {
 41         while (-1 != j && y[i] != x[j]) {
 42             sum = (sum+((1+j)*j/2)%mod)%mod;
 43             j = nxt[j];
 44         }
 45         i++; j++;
 46         //sum=(sum+j)%mod;
 47         if (j >= m)
 48         {
 49             sum = (sum+((1+j)*j/2)%mod)%mod;
 50             j = nxt[j];
 51         }
 52     }
 53     while(j>=1){
 54     	sum = (sum+((1+j)*j/2)%mod)%mod;
 55     	j=nxt[j];
 56     }
 57 }
 58 int main()
 59 {
 60    // clock_t startTime,endTime;  
 61     //startTime = clock();
 62     //freopen("data.in","r",stdin);
 63     //freopen("out.txt","w",stdout);
 64     int t;
 65     scanf("%d",&t);
 66     while(t--){
 67         scanf("%s%s",y,x);
 68         int len1=strlen(y);
 69         int len2=strlen(x);
 70         reverse(x,x+len2);
 71         reverse(y,y+len1);
 72         sum=0;
 73         KMP_Count(x,len2,y,len1);
 74         printf("%I64d\n",sum);
 75     }
 76     //endTime = clock();  
 77     //cout << "Totle Time : " <<(double)(endTime - startTime)<< "ms" << endl;
 78     return 0;
 79 }

View Code