hdu 1358 & hdu 3746 & poj 2406 & uva 12012 循环节与kmp
来源:互联网 发布:平板 win10 知乎 2017 编辑:程序博客网 时间:2024/06/16 13:07
参考
kmp next函数 kmp的周期问题,深入了解kmp中next的原理 ——Because Of You
HDU 1358
题意
对于给定的字符串
思路
充要条件:
Code
#include <bits/stdc++.h>#define maxn 1000010using namespace std;int f[maxn], n;char s[maxn];typedef long long LL;void getfail(char* P) { int m = strlen(P); f[0] = f[1] = 0; for (int i = 1; i < m; ++i) { int j = f[i]; while (j && P[j] != P[i]) j = f[j]; f[i+1] = P[i] == P[j] ? j+1 : 0; }}int kas;void work() { scanf("%s", s); getfail(s); printf("Test case #%d\n", ++kas); for (int i = 2; i <= n; ++i) { if (f[i] && i % (i-f[i]) == 0) printf("%d %d\n", i, i / (i-f[i])); } printf("\n");}int main() { while (scanf("%d", &n) != EOF && n) work(); return 0;}
HDU 3746
题意
对给定的串
思路
若本身就是(判定条件同上题),则补
否则找到开头处最小的(可能的)循环节,即
Code
#include <bits/stdc++.h>#define maxn 100010using namespace std;char s[maxn];int f[maxn];typedef long long LL;void getfail() { int n = strlen(s); f[0] = f[1] = 0; for (int i = 1; i < n; ++i) { int j = f[i]; while (j && s[i] != s[j]) j = f[j]; f[i+1] = s[i] == s[j] ? j+1 : 0; }}void work() { scanf("%s", s); getfail(); int n = strlen(s); int diff = n - f[n]; if (n % diff == 0 && n / diff > 1) printf("0\n"); else printf("%d\n", diff - n % diff);}int main() { int T; scanf("%d", &T); while (T--) work(); return 0;}
POJ 2406
题意
对于给定的字符串
思路
注意一下,如果不整除那么答案就是
Code
#include <cstdio>#include <cstring>#define maxn 1000010using namespace std;typedef long long LL;char s[maxn];int f[maxn], n;void getfail() { f[0] = f[1] = 0; for (int i = 1; i < n; ++i) { int j = f[i]; while (j && s[i] != s[j]) j = f[j]; f[i+1] = s[i] == s[j] ? j+1 : 0; }}void work() { n = strlen(s); getfail(); if (n % (n - f[n]) == 0) printf("%d\n", n / (n - f[n])); else printf("1\n");}int main() { while (scanf("%s", s) != EOF && s[0] != '.') work(); return 0;}
UVa 12012
题意
对于给定的字符串
思路
枚举起点,对于每一个起点开始的子串求一次
另外,若一个串能表示成
Code
#include <bits/stdc++.h>#define maxn 1010using namespace std;typedef long long LL;char s[maxn];int f[maxn], a[maxn], m, kas;void getfail(char* P, int l) { f[0] = f[1] = 0; for (int i = 1; i < l; ++i) { int j = f[i]; while (j && P[i] != P[j]) j = f[j]; f[i+1] = P[i] == P[j] ? j+1 : 0; }}void work() { memset(a, 0, sizeof(a)); scanf("%s", s); m = strlen(s); for (int i = 0; i < m; ++i) { getfail(s+i, m-i); for (int j = 1; j <= m-i; ++j) { int circ = j - f[j]; if (j % circ) continue; int cnt = j / circ; for (int k = 1; k * k <= cnt; ++k) { if (cnt % k == 0) a[cnt / k] = max(a[cnt / k], j), a[k] = max(a[k], j); } } } a[1] = m; printf("Case #%d:", ++kas); for (int i = 1; i <= m; ++i) printf(" %d", a[i]); printf("\n");}int main() { int T; scanf("%d", &T); while (T--) work(); return 0;}
阅读全文
0 0
- hdu 1358 & hdu 3746 & poj 2406 & uva 12012 循环节与kmp
- HDU 3746 kmp循环节
- hdu 1358 KMP循环节
- 【KMP(循环节)】poj 2406 Power Strings(外:hdu 1358 Period)
- 【KMP思想求循环节】hdu 1358 hust 1010 poj 2406
- [KMP-循环节问题]HDU 1358 period
- hdu 1358 period 求循环节 KMP
- hdu 3746(KMP) 最小循环节
- HDU 3746利用KMP找循环节
- hdu 1358 Period(KMP 循环节)
- HDU 1358 Period(kmp循环节)
- HDU 3746 Cyclic Nacklace (KMP 循环节)
- HDU 3746 KMP求循环节
- 【kmp循环节】hdu 3746 Cyclic Nacklace
- HDU - 3746 Cyclic Nacklace(KMP 循环节)
- hdu 1358 Period(KMP循环节)
- hdu 3746 kmp求字符串循环节
- HDU 3746 Cyclic Nacklace [KMP+循环节]
- cookie管理(1)
- Androidg 改变CheckBox下的选中与未选中图片
- 通达OA2017最新版20170905程序加密锁无限制直供
- Html中的meta详解
- 【mongodb】mongodb中的skip指的是什么
- hdu 1358 & hdu 3746 & poj 2406 & uva 12012 循环节与kmp
- SpringBoot学习笔记之JSP与freemarker支持
- android图表库MPAndroidChart之分组柱状图的里面的坑
- equals()与hashCode()
- SQL Server修改字段属性总结
- Python 直接赋值、浅拷贝和深度拷贝解析
- centOS7下的MySQL编码设置
- 【树形DP】宝藏
- Babel初体验