BestCoder Round #74 (div.2)【1排列组合】

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LCP Array

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1527 Accepted Submission(s): 465

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

330 043 2 131 2

Sample Output

16250260

大意：

问题描述

Peter有一个字符串 $s=s_{1}s_{2}...s_{n}$ , 令 $\text{suff}_i =s_{i}s_{i+1}...s_{n}$ 是 $s$ 第 $i$ 字符开头的后缀. Peter知道任意两个相邻的后缀的最长公共前缀 $a_i = \text{lcp}(\text{suff}_i, \text{suff}_{i+1}) \quad (1 \le i < n$ ).现在给你数组 $a$ , Peter有多少个仅包含小写字母的字符串满足这个数组. 答案也许会很大, 你只要输出对 $10^9 + 7$ 取模的结果即可.

输入描述

输入包含多组数据. 第一行有一个整数 $T$ , 表示测试数据的组数. 对于每组数据:第一行包含一个整数 $n$  ( $\le n \le 10^5)$ 表示字符串的长度. 第二行包含 $n - 1$ 个整数:  $a_1,a_2,...,a_{n-1}$   $\le a_i \le n)$ .所有数据中 $n$ 的和不超过 $10^6$ .

输出描述

对于每组数据, 输出答案对 $10^9+7$ 取模的结果.

思路：

观察 a[i] 可以发现，a[i] 是逐一递减的而且每次递减值为 1，直至递减为 0，再开始又一次的递减。

发现是排列组合问题等价于排列 26 种颜色，两两相邻不相同放同一种颜色，第一次肯定是放 26 种颜色啦，然后碰见一次 0 放一种颜色

#include<cstdio>#include<cstring>#include<algorithm>#define LL long longusing namespace std;const LL mod=1e9+7;int n;int a[100010];int main(){int t;scanf("%d",&t);while(t--){scanf("%d",&n);bool flag=1;LL ans=26;for(int i=1;i<n;i++){scanf("%d",a+i);if(a[i]>=n-i+1)flag=0;if(a[i-1]!=0&&a[i]!=a[i-1]-1){flag=0;}if(a[i]==0)ans=ans*25%mod;}if(flag)printf("%lld\n",ans);elseputs("0");}return 0;}

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