标签：let iostream other include hdu3336 ber ret code style
Count the string
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9588 Accepted Submission(s): 4480
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.
Output
For each case, output only one number: the sum of the match times for all the prefixes of s mod 10007.
Sample Input
Sample Output
Author
foreverlin@HNU
题意：
求长度为n的字符串中每个前缀在这条字符串中出现的次数的总和
代码：
//dp[i]表示字符串s[1~i]中出现了多少个以s[i]为结尾的前缀,又有f数组的特性，如果
//f[i]=j则s[1~j]=s[i-j+1~i]得出状态转移方程dp[i]=dp[f[i]]+1;
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mod=10007;
char P[200005];
int f[200005],dp[200005];
int main()
{
int t,n;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
scanf("%s",P);
f[0]=f[1]=0;
for(int i=1;i<n;i++){
int j=f[i];
while(j&&P[i]!=P[j]) j=f[j];
f[i+1]=(P[i]==P[j]?j+1:0);
}
dp[0]=0;
int sum=0;
for(int i=1;i<=n;i++){
dp[i]=dp[f[i]]+1;
sum+=dp[i]%mod;
sum%=mod;
}
printf("%d\n",sum);
}
return 0;
}
HDU3336 KMP+DP
标签：let iostream other include hdu3336 ber ret code style
原文地址：http://www.cnblogs.com/--ZHIYUAN/p/6420647.html
