hdu5635 BestCoder Round #74 (div.2) LCP Array

LCP Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 895    Accepted Submission(s): 252

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
   
   
   
   
   
   
   
   


   
   
   
   

首先每一个元素都限制在a[i] <= n - i， 若有超出限制的直接跳出输出0

发现规律a[i] != 0时a[i + 1] = a[i] - 1，若符合执行下一步，否则跳出，然后输出0

发现规律，有x个0，最后结果就是26 * 25^x

注意取模运算

#include <cstdio>
#include <iostream>

using namespace std;

int a[100005];
long long mod = 1000000007;

long long p(long long x) {
	long long res = 1;
	for (int i = 1; i <= x; i++) {
		res = (res * 25) % mod;
	}
	return res;
}

int main()
{
	int T, n;
	scanf("%d", &T);
	while (T--) {
		scanf("%d", &n);
		bool flag = true;
		int cou = 0;
		for (int i = 1; i <= n - 1; i++) {
			scanf("%d", &a[i]);
			if (!a[i]) cou++;
			if (a[i] > n - i) flag = false;
		}
		if (!flag) {
			puts("0");
			continue;
		}
		for (int i = 1; i <= n - 2; i++) {
			if (a[i] != 0 && a[i + 1] != a[i] - 1) {
				flag = false;
				break;
			}
		}
		if (flag) {
			printf("%I64d\n", (26 * p(cou)) % mod);
		}
		else {
			puts("0");
		}
	}
	return 0;
}