A Secret
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 256000/256000 K (Java/Others)
Total Submission(s): 3407 Accepted Submission(s): 1228
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
Recommend
liuyiding
简单的说，就是给你2个字符串，a,b，让你求 b满足Suffix(S2,i)规律的子串 在 a中出现的次数 和 子串本身长度 的乘积，然后累加。
爆搜显然是不行，最开始用了KMP，但是由于不是特别熟悉而且搞了n^2的复杂度，姿势不太好，无限TLE。后来用了扩展KMP，但是由于不知道数组开小了，TLE了2小时，年老眼花了。
#include<iostream>
#include<cstring>
#include<algorithm>
using namespace std;
#define mod 1000000007
const int maxn = 1000010;
long long Next[maxn], ex[maxn];
char a[maxn], b[maxn];
void GETNext(char *str)
{
int i = 0, j, po, len = strlen(str);
Next[0] = len;
while (str[i] == str[i + 1] && i + 1<len) i++;
Next[1] = i;
po = 1;
for (i = 2; i<len; i++)
{
if (Next[i - po] + i<Next[po] + po)
Next[i] = Next[i - po];
else
{
j = Next[po] + po - i;
if (j<0)j = 0;
while (i + j<len&&str[j] == str[j + i])
j++;
Next[i] = j;
po = i;
}
}
}
void EXKMP(char *s1, char *s2)
{
int i = 0, j, po, len = strlen(s1), l2 = strlen(s2);
GETNext(s2);
while (s1[i] == s2[i] && i<l2&&i<len)
i++;
ex[0] = i;
po = 0;
for (i = 1; i<len; i++)
{
if (Next[i - po] + i<ex[po] + po)
ex[i] = Next[i - po];
else
{
j = ex[po] + po - i;
if (j<0)j = 0;
while (i + j<len && j<l2 &&s1[j + i] == s2[j])
j++;
ex[i] = j;
po = i;
}
}
}
char *revstr(char *str, size_t len)
{
char *start = str;
char *end = str + len - 1;
char ch;
if (str != NULL)
{
while (start < end)
{
ch = *start;
*start++ = *end;
*end-- = ch;
}
}
return str;
}
int main()
{
int t;
scanf("%d", &t);
while (t--)
{
for (int i = 0; i < maxn; i++)
a[i] = b[i] = Next[i] = ex[i] = 0;
scanf("%s %s", a, b);
int len2 = strlen(b);
int len = strlen(a);
revstr(a, len); //因为扩展KMP求得是最大前缀长度，而我们要用的是后缀，所以翻转
revstr(b, len2);
EXKMP(a, b);
long long ans = 0;
for (int i = 0; i < len; i++)
{
ans = (ex[i] * (ex[i] + 1) / 2 % mod + ans) % mod;
}
printf("%lld\n", ans);
}
}