LCP Array

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 1564    Accepted Submission(s): 480 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 题意：是给出一个小写字母组成的串，从第一个字母开头到n-1个字母开头的子串 给出它与下一个子串的最长公共前缀，问该串的组成方式的种类

思路：我们可以观察到如果子串a[i]的公共前缀!=0,那么子串a[i+1]的公共前缀一定减1；而且前缀必须满足a[i]<=n-i

code

//#include <iostream> #include<cstdio> #define mod 1000000007 using namespace std; long long a[1000010],ans; int main() { int t,n; scanf("%d",&t); while(t--) { scanf("%d",&n); int flag=0; for(int i=1;i<n;i++) { scanf("%lld",&a[i]); if(flag) continue; if((a[i-1]!=0&&a[i]!=a[i-1]-1)||a[i]>n-i) flag=1; } if(flag) printf("0\n"); else{ ans=26; for(int i=1;i<n;i++) if(a[i]==0) ans=(ans*25)%mod; printf("%lld\n",ans); } } return 0; }