HDU 3336 Count the string
阿新 • • 發佈：2017-09-28
inpu sum problems eof monthly pro chmod ext cas
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. Input The 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.
Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11476 Accepted Submission(s):
5340
Problem Description 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. Input The 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.
#include<cstdio> #include<cstring> using namespace std; int n,m; char s[200010]; int next[200010],dp[200010]; void getnext(){ int k=-1,j=0; next[0]=-1; while(j<m){ if(k==-1||s[j]==s[k]){ j++;k++; next[j]=k; } else k=next[k]; } } int main(){ int T; scanf("%d",&T); while(T--){ memset(dp,0,sizeof(dp)); memset(next,0,sizeof(next)); scanf("%d",&m); scanf("%s",s); getnext(); dp[0]=0; int sum=0; for(int i=1;i<=m;i++){ dp[i]=dp[next[i]]+1; sum=(sum+dp[i])%10007; } printf("%d\n",sum); } }
HDU 3336 Count the string