【HDU】5635LCP Array（读题）

最新推荐文章于 2020-03-08 17:57:44 发布

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本文解析了一道涉及字符串匹配的算法题目，重点在于理解题意及如何利用前缀与后缀的最长公共前缀长度来确定可能的字符串数量。通过具体实例，文章详细阐述了算法思路与实现过程。

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1827 Accepted Submission(s): 588

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)(1≤i<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 (2≤n≤105) -- the length of the string. The second line contains n−1 integers: a1,a2,...,an−1 (0≤ai≤n) .

The sum of values of n in all test cases doesn't exceed 106 .

Output

For each test case output one integer denoting the answer. The answer must be printed modulo 109+7 .

Sample Input

3 3 0 0 4 3 2 1 3 1 2

Sample Output

16250 26 0

Source

BestCoder Round #74 (div.2)

Recommend

wange2014

题目大意：其实感觉这个题目主要就是把题目弄明白，当时就是没看懂题目是啥，

给你一个长度为N的字符串，给你前N-1个位置的a[i]，这个a[i]就是表示一个字符串以第i位开头的后缀与以第i+1位为开头的后缀的最长公共前缀的长度，其实我感觉就是整个N长度的字符串都是一个字母，比如是A，就是如果当前有一个a[i]为0，那么就是当前的这个字母不是A，而是其他的，问你这种字符串的数量有多少个

思路：就像上面说的，比如这个字母是A，那么这个A就有26种，‘a’~‘z’随意一个，然后后面说的不是A的就有25种，因为26-1嘛，写出来就好了，还是好好读题吧，

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int mod=1000000007;
const int maxn=100050;
int n;
int a[maxn];

int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1; i<=n-1; i++)
        {
            scanf("%d",&a[i]);
        }

       long long ans=26;
        for(int i=1; i<=n-1; i++)
        {
            if(a[i-1]!=0&&a[i]!=a[i-1]-1)
            {
                ans=0;
                break;
            }

            if(a[i]==0)
            {
                ans=(ans*25)%mod;
            }
        }
        if(a[n-1]>1)
        {
            cout<<0<<endl;
            continue;
        }
        cout<<ans<<endl;
    }
}