LCP Array

**Time limit:4000/2000 MS (java/others) Memory limit:131072/131072 K (java/others)**

Total submission (s): 830 Accepted Submission (s): 232

Problem Descriptionpeter has a strings =s1s 2 sn , letSuffI= Sis I+1 sn Being the suffix start withI-th character ofs. Peter knows the LCP (longest common prefix) of each of the adjacent suffixes which denotes asAI=Lcp(suffi,suffi+1)(1≤i<n ).

Given the LCP array, Peter wants to know how many strings containing lowercase 中文版 letters only would satisfy the LCP a Rray. The answer may too large, just print it modulo 9+7.

Inputthere is multiple test cases. The first line of input contains an integerTindicating the number of test cases. For each test case:

The first line contains an integerN(2≤n≤5) -The length of the string. The second line containsn−1 Integers:a1,a2,.. . ,an−1 (0≤ai≤n) .

The sum of values of n in all test cases doesn ' t exceed 6.

Outputfor each test case output one integer denoting the answer. The answer must be printed modulo9+7.

Sample Input330 0 43 2 1 31 2

Sample Output16250260

Sourcebestcoder Round #74 (Div.2) Topic: PROBLEM-solving ideas: Because the AI is the longest common prefix of the adjacent suffix LCP, then we handwritten a few samples corresponding to the string to see if there is any law, we found when a[i-1]!=0& &a[i] >= A[i-1] is not possible to exist the corresponding string such as 1 2 0 or 1 1 0, even if the decrement, should be reduced by 1, and A[i] <= n-i, that is suff[i] and suff[i+1] The longest public prefix should be less than n-i, There is a restrictive relationship. There is a 0 of the number is the result should be multiplied by the number of 25, if the given data is a non-0 solution, then the result of the minimum solution should be 26, so initialized to 26.

#include <stdio.h> #include <string.h> #include <iostream> #include <vector> #include < math.h> #include <map> #include <set> #include <queue> #include <stack> #include <string > #include <stdlib.h> #include <algorithm>using namespace std;typedef long long ll;const int maxn = 1e6; Const LL MOD = 1000000007;int a[maxn];int main () {int cas; scanf ("%d", &cas); while (cas--) {int n, flag = 0, c = 0; scanf ("%d", &n); for (int i = 1; i < n; i++) {scanf ("%d", &a[i]); if (!a[i]) C + +; if (A[i] > N-i) {//restriction flag = 1; } if (a[i-1]! = 0 && a[i-1]-a[i]! = 1) {//minus 1 decrements flag = 1; }} if (flag) {puts ("0"); continue; } LL ans = 26; for (int i = 1; I <= C; i++) {ans = (ans *)% mod; } printf ("%lld\n", ans); } return 0;} /*5550 1 0 4501 0 243 2 150 1 2 070 0 3 2 1 0*/

HDU 5635--LCP Array —————— "idea question"