HDU 4333 Revolving Digits 扩张KMP

意甲冠军：求一个数字环路移动少于不同数量等同于的数字

思路：扩展KMP求出S[i..j]等于S[0..j-i]的最长前缀推断 next[i] 大于等于n就是同样小于n推断S[next[i]]和S[next[i]+i]的大小

next数组的含义就是S字符串以i開始的和S本身(以0開始)的最长公共前缀把题目输入的复制一倍然后直接推断next数组比較大小即可了

#include <cstdio>
#include <cstring>
using namespace std;

const int maxn = 200010;
int f[maxn];
char a[maxn];

void getFail(char* s)
{
	int n = strlen(s);
	int p = 0, k = 1;
	for(int i = 0; i+1 < n && s[i] == s[i+1]; i++)
		p++;
	f[0] = n;	
	f[1] = p;
	for(int i = 2; i < n/2; i++)
	{
		int p = f[k]+k-1;
		int l = f[i-k];
		if(i+l-1 < p)
			f[i] = l;
		else
		{
			int j = p-i+1;
			if(j < 0)
				j = 0;
			while(i+j < n*2 && s[i+j] == s[j])
				j++;
			f[i] = j;
			k = i;
		}
	}
}
int main()
{
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		scanf("%s", a);
		int n = strlen(a);
		for(int i = 0; i < n; i++)
		{
			a[i+n] = a[i];
		}
		a[2*n] = 0;
		getFail(a);
		int len;
		for(len = 1; len <= n; len++)
		{
			if(f[len]+len >= n)
			{
				if(n%len)
					len = n;
				break;
			}
		}
		int sum1 = 0, sum2 = 0, sum3 = 0;
		for(int i = 0; i < len; i++)
		{
			if(f[i] >= n)
				sum2++;
			else if(a[i+f[i]] > a[f[i]])
				sum3++;
			else
				sum1++;
		}
		printf("Case %d: %d %d %d
", cas++, sum1, sum2, sum3);
	}
	return 0;
}