POJ 2406Power Strings(KMP)
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 }
