hdu 6153 A Secret KMP A Secret

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概述

A Secret

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

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

S，T先逆序处理后，相当于统计T中所有前缀t在S中出现的次数；

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;
public class Main {
public static void main(String[] args) {
new Task().solve();
}
}
class Task {
InputReader in = new InputReader(System.in) ;
PrintWriter out = new PrintWriter(System.out) ;
final long Mod = 1000000007L ;
final int maxn = 1000008 ;
int n,m;
char[] a = new char[maxn] ;
char[] b = new char[maxn] ;
int[] p = new int[maxn] ;
int[] dp = new int[maxn] ;
void getp(){
p[1]=0;
int i,j=0;
for(i=2;i<=m;i++){
while(j>0&&b[j+1]!=b[i]){
j=p[j];
}
if(b[j+1]==b[i]){
j+=1;
}
p[i]=j;
}
}
void kmp(int cnt[]){
int i,j=0;
for(i=1;i<=n;i++){
while(j>0&&b[j+1]!=a[i]){
j=p[j];
}
if(b[j+1]==a[i]){
j+=1;
cnt[j]++;
}
if(j==m){
j=p[j];
}
}
for(i=m;i>=1;i--){
if(p[i]!=0){
cnt[p[i]]+=cnt[i];
}
}
}
void reverse(char s[] , int len){
for(int i=1;i<=len/2;i++){
char t = s[i] ;
s[i] = s[len-i+1] ;
s[len-i+1] = t ;
}
}
void solve(){
int t = in.nextInt() ;
while(t-- > 0){
String s = in.next() ;
n = s.length() ;
for(int i = 1 ; i <= n ; i++){
a[i] = s.charAt(i-1) ;
}
s = in.next() ;
m = s.length() ;
for(int i = 1 ; i <= m ; i++){
b[i] = s.charAt(i-1) ;
}
reverse(a , n);
reverse(b , m);
Arrays.fill(dp, 0) ;
getp();
kmp(dp);
long sum = 0 ;
for(int i = 1 ; i <= m ; i++){
sum += ((long)dp[i] * i % Mod) ;
sum %= Mod ;
}
out.println(sum) ;
}
out.flush() ;
}
}
class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = new StringTokenizer("");
}
private void eat(String s) {
tokenizer = new StringTokenizer(s);
}
public String nextLine() {
try {
return reader.readLine();
} catch (Exception e) {
return null;
}
}
public boolean hasNext() {
while (!tokenizer.hasMoreTokens()) {
String s = nextLine();
if (s == null)
return false;
eat(s);
}
return true;
}
public String next() {
hasNext();
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public int[] nextInts(int n) {
int[] nums = new int[n];
for (int i = 0; i < n; i++) {
nums[i] = nextInt();
}
return nums;
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
}