hdu-5635 LCP Array

Posted LittlePointer

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了hdu-5635 LCP Array相关的知识,希望对你有一定的参考价值。

LCP Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 358    Accepted Submission(s): 102


Problem Description
Peter has a string s=s1s2...sn, let suffi=sisi+1...sn be the suffix start with i-th character of s. Peter knows the lcp (longest common prefix) of each two adjacent suffixes which denotes as ai=lcp(suffi,suffi+1)(1i<n).

Given the lcp array, Peter wants to know how many strings containing lowercase English letters only will satisfy the lcp array. The answer may be too large, just print it modulo 109+7.
 

 

Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer n (2n10^5) -- the length of the string. The second line contains n1 integers: a1,a2,...,an1 (0ain).

The sum of values of n in all test cases doesn‘t exceed 10^6.
 

 

Output
For each test case output one integer denoting the answer. The answer must be printed modulo 1e9+7.
 

 

Sample Input
3
3
0 0
4
3 2 1
3
1 2
 

 

Sample Output
16250
26
0
AC代码:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
const int N=1e5+5;
const long long mod=1e9+7;
int n,a[N];
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
scanf("%d",&n);
for(int i=1;i<n;i++)
{
scanf("%d",&a[i]);
}
a[n]=0;
int flag=1;
for(int i=1;i<n;i++)
{
if(a[i])
{
if(a[i]-a[i+1]!=1)
{
flag=0;
break;
}
}
}
if(!flag)
{
cout<<"0"<<"\n";
}
else
{
int num=0;
for(int i=1;i<n;i++)
{
if(!a[i])
{
num++;
}
}
long long ans=26;
for(int i=0;i<num;i++)
{
ans=(ans*25)%mod;
}
cout<<ans%mod<<"\n";
}

}
return 0;
}

 

以上是关于hdu-5635 LCP Array的主要内容,如果未能解决你的问题,请参考以下文章

扩展KMP(Z函数),线性LCP

后缀数组入门——Height数组与LCP

PHP中array_merge()函数与array+arrayarray_merge_recursive() 的区别

bzoj1692: [Usaco2007 Dec]队列变换(hash+二分求LCP)

LightHouse 和 Performance 之间的 LCP 时间 - Google Chrome

c_cpp 使用后缀数组查找最长公共前缀(LCP)。复杂性:SA = O(n.log(n)),LCP = O(n)