牛客网暑期ACM多校训练营(第三场): A. Ternary String（欧拉降幂+递推）

本文探讨了一种特殊的ternary字符串，在特定规则下自我复制直至最终可能变为空字符串的问题。通过对每一步变化规律的研究，利用数学方法求解该字符串完全消失所需的秒数，并通过模运算处理大数值问题。

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题目描述

A ternary string is a sequence of digits, where each digit is either 0, 1, or 2.
Chiaki has a ternary string s which can self-reproduce. Every second, a digit 0 is inserted after every 1 in the string, and then a digit 1 is inserted after every 2 in the string, and finally the first character will disappear.
For example, ``212'' will become ``11021'' after one second, and become ``01002110'' after another second.
Chiaki would like to know the number of seconds needed until the string become an empty string. As the answer could be very large, she only needs the answer modulo (109 + 7).

输入描述:

There are multiple test cases. The first line of input is an integer T indicates the number of test cases. For each test case:
The first line contains a ternary string s (1 ≤ |s| ≤ 105).
It is guaranteed that the sum of all |s| does not exceed 2 x 106.

输出描述:

For each test case, output an integer denoting the answer. If the string never becomes empty, output -1 instead.

输入

输出

3
93
45

有道处理方法一样的题：https://blog.youkuaiyun.com/Jaihk662/article/details/77927831

引用官方题解：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<map>
using namespace std;
#define LL long long
map<LL, LL> p;
char str[2000005];
LL Euler(LL x)
{
	LL i, ans;
	if(x==1)
		return 1;
	ans = x;
	for(i=2;i*i<=x;i++)
	{
		if(x%i==0)
			ans = ans/i*(i-1);
		while(x%i==0)
			x /= i;
	}
	if(x!=1)
		ans = ans/x*(x-1);
	return ans;
}
LL Pow(LL a, LL b, LL mod)
{
	LL ans;
	ans = 1;
	while(b)
	{
		if(b%2)
			ans = ans*a%mod;
		a = a*a%mod;
		b /= 2;
	}
	return ans;
}
LL Jud(LL loc, LL mod)
{
	if(loc==0)
		return 0;
	if(str[loc]=='0')
		return (Jud(loc-1, mod)+1)%mod;
	if(str[loc]=='1')
		return 2*(Jud(loc-1, mod)+1)%mod;
	return (3*Pow(2, Jud(loc-1, p[mod])+1, mod)+mod-3)%mod;
}
int main(void)
{
	LL T, now, n;
	now = 1000000007;
	p[1] = 1;
	while(now!=1)
	{
		p[now] = Euler(now);
		now = p[now];
	}
	scanf("%lld", &T);
	while(T--)
	{
		scanf("%s", str+1);
		n = strlen(str+1);
		printf("%lld\n", Jud(n, 1000000007));
	}
	return 0;
}