【HDU 5635 LCP Array】

LCP Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1554 Accepted Submission(s): 476

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)

统计不同的字符个数，若a[i - 1] 不为零，者a[i] 必为 a[i - 1] - 1 ，第i 个字符与地i- 1 个字符相同,若a[i - 1]为零，者 第 i 个字符与地i- 1 个字符不同

AC代码：

#include<cstdio>typedef long long LL;const int K = 1e5 + 10;const LL mod = 1e9 + 7;int a[K];int main(){    int T,N;    scanf("%d",&T);    while(T--){        scanf("%d",&N);        LL ans = 26,ok = 1;        for(int i = 1; i < N; i++){            scanf("%d",&a[i]);            if(!a[i]) ans = (ans * 25) % mod;            if(a[i - 1] && a[i] != a[i - 1] - 1 || a[i] > N - i) ok = 0;        }        if(ok) printf("%lld\n",ans);        else printf("0\n");    }    return 0;}