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(Incomplete) UVa 10298 Power Strings
方法:暴力 / kmp
先说一下暴力的方法,就是尝试把原来的string不断分割,在substring 里求解。code 如下
#include <cstdio> #include <cstring> #include <algorithm> #include <iostream> #include <string> #include <vector> #include <stack> #include <bitset> #include <cstdlib> #include <cmath> #include <set> #include <list> #include <deque> #include <map> #include <queue> #include <fstream> #include <cassert> #include <unordered_map> #include <cmath> #include <sstream> #include <time.h> #include <complex> #include <iomanip> #define Max(a,b) ((a)>(b)?(a):(b)) #define Min(a,b) ((a)<(b)?(a):(b)) #define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a)) #define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a)) #define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a)) #define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a)) #define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a)) #define FOREACH(a,b) for (auto &(a) : (b)) #define rep(i,n) FOR(i,0,n) #define repn(i,n) FORN(i,1,n) #define drep(i,n) DFOR(i,n-1,0) #define drepn(i,n) DFOR(i,n,1) #define MAX(a,b) a = Max(a,b) #define MIN(a,b) a = Min(a,b) #define SQR(x) ((LL)(x) * (x)) #define Reset(a,b) memset(a,b,sizeof(a)) #define fi first #define se second #define mp make_pair #define pb push_back #define all(v) v.begin(),v.end() #define ALLA(arr,sz) arr,arr+sz #define SIZE(v) (int)v.size() #define SORT(v) sort(all(v)) #define REVERSE(v) reverse(ALL(v)) #define SORTA(arr,sz) sort(ALLA(arr,sz)) #define REVERSEA(arr,sz) reverse(ALLA(arr,sz)) #define PERMUTE next_permutation #define TC(t) while(t--) #define forever for(;;) #define PINF 1000000000000 #define newline ‘\n‘ #define test if(1)if(0)cerr using namespace std; using namespace std; typedef vector<int> vi; typedef vector<vi> vvi; typedef pair<int,int> ii; typedef pair<double,double> dd; typedef pair<char,char> cc; typedef vector<ii> vii; typedef long long ll; typedef unsigned long long ull; typedef pair<ll, ll> l4; const double pi = acos(-1.0); int main() { ios::sync_with_stdio(false); cin.tie(0); string str; while (cin >> str) { if (str == ".") break; int n, m; n = m = str.length(); for (int i = 2; i <= n; ++i) { while (n % i == 0) { n /= i; int j = 0; int len = m/i; while (j < m-len && str[j] == str[j+len]) ++j; if (j == m-m/i) { m = len; } else break; } //cerr << i << " " << m << newline; while (n % i == 0) n /= i; } cout << str.length()/m << newline; } }
然而看到题解,好像是用kmp,值得学习
http://www.cnblogs.com/jackge/archive/2013/01/05/2846006.html
(Incomplete) UVa 10298 Power Strings
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