【poj2406】Power Strings——KMP
阿新 • • 發佈:2018-12-24
題目:
Power StringsTime Limit: 3000MS | Memory Limit: 65536K |
Total Submissions: 41220 | Accepted: 17140 |
Description
Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).Input
Output
For each s you should print the largest n such that s = a^n for some string a.Sample Input
abcd aaaa ababab .
Sample Output
1 4 3
Hint
This problem has huge input, use scanf instead of cin to avoid time limit exceed.Source
描述:求字串s的迴圈節個數
題解:
if (len % (len - nxt[len]) == 0) // 看整個長度len能分解成x個這種串。
ans = len / (len - nxt[len]);
程式碼:
#include <cstdio> #include <cstring> #include <algorithm> #include <cstdlib> using namespace std; const int maxn = 1000005; char text[maxn]; int nxt[maxn]; int main() { while (scanf("%s", text) != EOF) { if (text[0] == '.')break; int len = strlen(text); for (int i = 0, j = -1; i <= len; i++, j++) { nxt[i] = j; while (~j && text[i] != text[j]) j = nxt[j]; } int ans = 1; if (len % (len - nxt[len]) == 0) // 看整個長度len能分解成x個這種串。 ans = len / (len - nxt[len]); printf("%d\n", ans); } return 0; }