It is well known that AekdyCoin is good at string problems as well as number theory problems. When given a string s, we can write down all the non-empty prefixes of this string. For example:
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can
see that prefix "a" matches twice, "ab" matches twice too, "aba" matches
once, and "abab" matches once. Now you are asked to calculate the sum
of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1
= 6.
The answer may be very large, so output the answer mod 10007.
s: "abab"
The prefixes are: "a", "ab", "aba", "abab"
For each prefix, we can count the times it matches in s. So we can
see that prefix "a" matches twice, "ab" matches twice too, "aba" matches
once, and "abab" matches once. Now you are asked to calculate the sum
of the match times for all the prefixes. For "abab", it is 2 + 2 + 1 + 1
= 6.
The answer may be very large, so output the answer mod 10007.
InputThe first line is a single integer T, indicating the number of test cases.
For each case, the first line is an integer n (1 <= n <=
200000), which is the length of string s. A line follows giving the
string s. The characters in the strings are all lower-case letters.
OutputFor each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.Sample Input
For each case, the first line is an integer n (1 <= n <=
200000), which is the length of string s. A line follows giving the
string s. The characters in the strings are all lower-case letters.
OutputFor each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.Sample Input
1
4
abab
Sample Output
6 kmp预处理Next数组。然后遍历每一位。dp[i]=dp[Next[i]]+1 即可
#include<stdio.h>
const int maxn=;
#define mod 10007
int _,n,Next[maxn],d[maxn];
char s[maxn]; void prekmp() {
int i,j;
j=Next[]=-;
i=;
while(i<n) {
while(j!=-&&s[i]!=s[j]) j=Next[j];
Next[++i]=++j;
}
} int main() {
for(scanf("%d",&_);_;_--) {
scanf("%d%s",&n,s);
prekmp();
d[]=;
int sum=;
for(int i=;i<=n;i++) {
d[i]=d[Next[i]]+;
sum+=d[i]%mod;
}
printf("%d\n",sum%mod);
}
}