正题

题目链接:https://www.luogu.com.cn/problem/P4884

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题目大意

求一个最小的$n$使得$n$个连续的$1$其在模$m$意义下等于$k$。

$6\le m\le 10^{11},0<k<m$

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解题思路

补一道老题

$n$个连续的$1$就是$\frac{10^n-1}{9}$所以题目是求
$$\frac{10^n-1}{9}\equiv k(modm)$$
$$\Rightarrow 10^n\equiv 9k+1(modm)$$
然后$\text{BSGS}$就好了

要开$\text{\_\_int128}$

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code

#include<cstdio>
#include<cmath>
#include<map>
#include<cctype>
#define ll __int128
using namespace std;
ll read(){ll x=0,f=1;char c=getchar();while(!isdigit(c)){if(c=='-')f=-f;c=getchar();}while(isdigit(c)){x=(x<<1)+(x<<3)+c-'0';c=getchar();}return x*f;
}
void print(ll x)
{if(x>9)print(x/10);putchar(48+x%10);return;}
ll k,m,a;
map<ll,ll> hash;
int main()
{k=read();m=read();k=(k*9+1)%m;ll q=(ll)sqrt((double)m)+1;ll val=1;for(ll j=0;j<q;j++){hash[val*k%m]=j;val=val*10%m;}a=val%m;if(!a) {printf("%d",(m==0)?1:-1);return 0;}val=1;for(ll i=0;i<=q;i++){ll j=hash.find(val)==hash.end()?-1:hash[val];if(j>=0&&i*q-j>=0) {print(i*q-j);return 0;}val=val*a%m;}printf("-1");
}