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最新推荐文章于 2016-07-12 19:06:23 发布

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LCP Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 1223 Accepted Submission(s): 338

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

Sample Output

Source

BestCoder Round #74 (div.2)

//主要符合a[i]=a[i-1]-1 && a[i]!=0 a[i]<=n-i-1

#include <bits/stdc++.h>
using namespace std;
__int64 Mod = 1000000007;
int a[100005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,flag=0;
        scanf("%d",&n);
        for(int i=0;i<n-1;i++)
        {
            scanf("%d",a+i);
            if(i>0&&a[i]!=a[i-1]-1&&a[i-1]!=0)
                flag=1;
            if(a[i]>n-i-1)
                flag=1;
        }
        if(flag)
            puts("0");
        else
        {
             __int64 sum=26;
             for(int i=0;i<n-1;i++)
             {
                  if(a[i]==0)
                     sum=(sum*25)%Mod;
             }
                printf("%I64d\n",sum);
        }

    }

    return 0;
}