hdoj 5635 LCP Array (模拟)
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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 modulo109+7 .
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
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 integern (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 ofn in all test cases doesn't exceed 106 .
The first line contains an integer
The sum of values of
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
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 1000000007using 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;}
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