POJ 2406Power Strings(KMP)

POJ 2406

其实就是一个简单的kmp应用：

ans = n % (n - f[n]) == 0 ? n / (n - f[n]) : 1

其中f是失配函数

 1 //#pragma comment(linker, "/STACK:1677721600")
 2 #include <map>
 3 #include <set>
 4 #include <stack>
 5 #include <queue>
 6 #include <cmath>
 7 #include <ctime>
 8 #include <vector>
 9 #include <cstdio>
10 #include <cctype>
11 #include <cstring>
12 #include <cstdlib>
13 #include <iostream>
14 #include <algorithm>
15 using namespace std;
16 #define INF 0x3f3f3f3f
17 #define inf (-((LL)1<<40))
18 #define lson k<<1, L, (L + R)>>1
19 #define rson k<<1|1,  ((L + R)>>1) + 1, R
20 #define mem0(a) memset(a,0,sizeof(a))
21 #define mem1(a) memset(a,-1,sizeof(a))
22 #define mem(a, b) memset(a, b, sizeof(a))
23 #define FIN freopen("in.txt", "r", stdin)
24 #define FOUT freopen("out.txt", "w", stdout)
25 #define rep(i, a, b) for(int i = a; i <= b; i ++)
26 #define dec(i, a, b) for(int i = a; i >= b; i --)
27
28 template<class T> T MAX(T a, T b) { return a > b ? a : b; }
29 template<class T> T MIN(T a, T b) { return a < b ? a : b; }
30 template<class T> T GCD(T a, T b) { return b ? GCD(b, a%b) : a; }
31 template<class T> T LCM(T a, T b) { return a / GCD(a,b) * b;    }
32
33 //typedef __int64 LL;
34 typedef long long LL;
35 const int MAXN = 1000000 + 10;
36 const int MAXM = 110000;
37 const double eps = 1e-8;
38 LL MOD = 1000000007;
39
40 char s[MAXN];
41 int f[MAXN];
42
43 void get_next(int n) {
44     mem0(f);
45     rep (i, 1, n - 1) {
46         int j = f[i];
47         while(j && s[i] != s[j])  j = f[j];
48         f[i + 1] = s[i] == s[j] ? j + 1 : 0;
49     }
50 }
51
52 int main()
53 {
54     //FIN;
55     while(~scanf("%s", s) && s[0] != ‘.‘) {
56         int n = strlen(s);
57         get_next(n);
58         printf("%d\n", n % (n - f[n]) == 0 ? n / (n - f[n]) : 1);
59     }
60     return 0;
61 }