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原创最新推荐文章于 2026-06-23 16:26:19 发布 656 阅读 AI 写同款· GEO 优化 ›

M365 Copilot在Power Apps中的Vibe Coding实战指南 Vibe Coding 是一种基于自然语言驱动低代码开发的新型人机协同范式，其核心在于将业务意图实时翻译为 Power Apps 可执行的控件结构、Power Fx 公式与事件逻辑。它依托 Microsoft 365 Copilot 的语义理解能力，在 App Builder 场景中实现设计时（Design-time）的智能辅助，而非运行时自动化。该技术显著降低原型构建门槛，提升从业务需求到可交互界面的转化效率，广泛应用于一人团队项目开发、销售看板搭建、员工自助服务等轻量级业务应用快速交付场景。本文聚焦 M 阅读详情

Time Limit: 3000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

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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

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

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 li

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
#define MAXN 1000010 
int next[MAXN];
int len;
char s[MAXN];

void get_next ()
{
	int i=0,j=-1;
	next[0]=-1;
	while(i<len)
	{
		if(j==-1||s[i]==s[j])
		{
			i++;
			j++;
			next[i]=j;
		}
		else j=next[j];
	}
}

int main()
{
	while(scanf("%s",s)!=EOF)
	{
		if(s[0]=='.')break;
		len =strlen(s);
		int k;
		get_next();
		if(len%(len-next[len])==0)

  /*
        因为next[i]存储的是s[i]为结尾的字符串的最大前缀
        而本题要求子串相连不能相交,所以为了避免相交,需加以判断 
        如果能够整除则输出商 否则输出1 
        */
		{
			k=len/(len-next[len]);
			printf("%d\n",k);
		}
		else printf("1\n");
	}
	return 0;
}