CodeForces 432D|Prefixes and Suffixes|KMP|动态规划
2016-11-18 00:35
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题目大意
求s的不同长度的既是s前缀又是s后缀的子串的个数题解
考虑KMP的next数组。因为next[j]表示lcp(s[1..j],s[1..k])(k<j)。
故s[1..j]与s[1..next[j]]的关系是后者是前者的最长后缀,由于是最长,所以肯定不会漏情况,因此后者的出现次数是前者的出现次数+1(前者最后一次出现的后缀)。。。。。。。。。
所以如果从s[1..n]开始,那么s[1..next
]一定是字符串的前后缀,s[1..next[next
]一定是s[1..next
]的前后缀,也肯定是字符串的前后缀,以此类推,然后DP方程就好说了。
dp[next[i]] += dp[i]
最后还有dp[i]+1表示最后一次的出现。
#include <cstdio> #include <cstring> const int N = 100005; char p ; int next , dp , len ; int main() { int i, j, n, tot = 0; scanf("%s", p + 1); n = strlen(p + 1); for (i = 0, j = 2; j <= n; ++j) { while (i && p[j] != p[i + 1]) i = next[i]; if (p[i + 1] == p[j]) ++i; next[j] = i; } for (i = n; i; i = next[i]) len[++tot] = i; for (i = n; i; --i) dp[next[i]] += ++dp[i]; printf("%d\n", tot); for (i = tot; i; --i) printf("%d %d\n", len[i], dp[len[i]]); return 0; }
D. Prefixes and Suffixes
Description
You have a string s = s1s2…s|s|, where |s| is the length of string s, and si its i-th character.Let’s introduce several definitions:
A substring s[i..j] (1 ≤ i ≤ j ≤ |s|) of string s is string sisi + 1…sj.
The prefix of string s of length l (1 ≤ l ≤ |s|) is string s[1..l].
The suffix of string s of length l (1 ≤ l ≤ |s|) is string s[|s| - l + 1..|s|].
Your task is, for any prefix of string s which matches a suffix of string s, print the number of times it occurs in string s as a substring.
Input
The single line contains a sequence of characters s1s2…s|s| (1 ≤ |s| ≤ 105) — string s. The string only consists of uppercase English letters.Output
In the first line, print integer k (0 ≤ k ≤ |s|) — the number of prefixes that match a suffix of string s. Next print k lines, in each line print two integers li ci. Numbers li ci mean that the prefix of the length li matches the suffix of length li and occurs in string s as a substring ci times. Print pairs li ci in the order of increasing li.Examples
Input
ABACABAOutput
31 4
3 2
7 1
Input
AAAOutput
31 3
2 2
3 1
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