HDU 5635 LCP array

LCP Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 307 Accepted Submission(s): 86

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

题目就是说给你输入的a[i]后 第i个字符与后面a[i]个字符都是相同的 问你有多少个符合输入的种类
本题有主要有三个点
1.a[i]小于n-i
2.a[i]是每一段递减的 而且必须是依次减1 减到0后可以接着再有一段递减的 但最后都要到1 否则不符合1
3.每一段递减段的字符必须是相同的
可以把每一段规程一块 且相邻的两块不相同

Ps：取鱼之后的输出也超了int 要用long long int
博主在这里错了无数发才想到这个问题
最后正常写就好了

#include <stdio.h>
const int MOD=1000000007;
int a[100005];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        long long int sum=26,mm=0;
        for(int i=1; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        for(int i=1; i<n; i++)
        {
            if(a[i]>n-i||a[i-1]!=0&&a[i]!=a[i-1]-1)
            {
                sum=0;
                break;
            }
            if(a[i]==0)
            {
                sum*=25;
                sum%=MOD;
            }
        }
        printf("%I64d\n",sum%MOD);
    }
    return 0;
}