hdu 5365

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

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

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

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)

//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;
}