hdu 5635 LCP Array（BC第一题）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5635

LCP Array

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

Total Submission(s): 282    Accepted Submission(s): 79

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) 

题目大意：先输入一个t代表着測试组数。再输入一个n表示字符串的长度。s1,s2,s2,s2.......接下去输入n-1个数，表示的最长同样前缀的长度。

即：以第二组測试数据为例

 ai=lcp(suffi,suffi+1);

解题思路：经过分析找到：a[i-1]=a[i]+1。假设一直满足这个条件的话，直接输出26，不满足直接输出0。

须要注意的是：最后一个数a[n-1]一定要大于1。否则直接输出0。

另一个就是第一数a[1]等于0的话，直接乘26。之后遇到0就乘25。

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

__int64 a[100010],s;
const int Mod=(1e9+7);

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        int n;
        scanf("%d",&n);
        for (int i=1; i<=n-1; i++)
        {
            scanf ("%I64d",&a[i]);
        }
        int flag=0;
        s=1;
        int k=0;
        for (int i=1; i<n; i++)
        {
            if (i==1)
            {
                if (a[i]==0)
                {
                    s=(s*26)%Mod;
                    k=1;
                }
                continue;
            }
            if (a[i-1]>0&&a[i-1]!=a[i]+1)
                flag=1;
            else
            {
                if (a[i]==0)
                {
                    if (k==1)
                        s=(s*25)%Mod;
                    else
                    {
                        s=(s*26)%Mod;
                        k=1;
                    }
                }
            }
        }
        if (k==1)
            s=(s*25)%Mod;
        else
            s=(s*26)%Mod;
        if (flag==1||a[n-1]>1)
            printf ("0
");
        else
            printf ("%I64d
",s);
    }
    return 0;
}

ai=lcp(suffi,suffi+1)