LuoguP4884题解

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BSGS模板 +　小学奥数技巧

题目描述

给定整数$K$和质数$m$，求最小的正整数$N$，使得$11…1(\text{（N个1）}) \equiv K(mod ;m)$．

Input & Output’s Examples

Inputs’ e.g. #1

1
9 17

Outputs’ e.g.#1

1
3

分析

这题乍一看不好解决，所以说我们需要一点小学奥数的技巧．

首先我们把同余式两边乘上一个$9$,再加上一个$1$,就可以得到：

$$10^N \equiv 9K + 1(mod ; m)$$

然后我们直接套用BSGS模板求解即可,注意，在进行BSGS的时候可能会两个long long相乘而爆掉，于是我们需要手写快速乘．

代码

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// luogu-judger-enable-o2
/* Headers */
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cctype>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<climits>
#include<iostream>
#include<map>
#define FOR(i,a,b,c) for(int i=(a);i<=(b);i+=(c))
#define ROF(i,a,b,c) for(int i=(a);i>=(b);i-=(c))
#define FORL(i,a,b,c) for(long long i=(a);i<=(b);i+=(c))
#define ROFL(i,a,b,c) for(long long i=(a);i>=(b);i-=(c))
#define FORR(i,a,b,c) for(register int i=(a);i<=(b);i+=(c))
#define ROFR(i,a,b,c) for(register int i=(a);i>=(b);i-=(c))
#define lowbit(x) x&(-x)
#define LeftChild(x) x<<1
#define RightChild(x) (x<<1)+1
#define RevEdge(x) x^1
#define FILE_IN(x) freopen(x,"r",stdin);
#define FILE_OUT(x) freopen(x,"w",stdout);
#define CLOSE_IN() fclose(stdin);
#define CLOSE_OUT() fclose(stdout);
#define IOS(x) std::ios::sync_with_stdio(x)
#define Dividing() printf("-----------------------------------\n");
namespace FastIO{
    const int BUFSIZE = 1 << 20;
    char ibuf[BUFSIZE],*is = ibuf,*its = ibuf;
    char obuf[BUFSIZE],*os = obuf,*ot = obuf + BUFSIZE;
    inline char getch(){
        if(is == its)
            its = (is = ibuf)+fread(ibuf,1,BUFSIZE,stdin);
        return (is == its)?EOF:*is++;
    }
    inline int getint(){
        int res = 0,neg = 0,ch = getch();
        while(!(isdigit(ch) || ch == '-') && ch != EOF)
            ch = getch();
        if(ch == '-'){
            neg = 1;ch = getch();
        }
        while(isdigit(ch)){
            res = (res << 3) + (res << 1)+ (ch - '0');
            ch = getch();
        }
        return neg?-res:res;
    }
    inline void flush(){
        fwrite(obuf,1,os-obuf,stdout);
        os = obuf;
    }
    inline void putch(char ch){
        *os++ = ch;
        if(os == ot)	flush();
    }
    inline void putint(int res){
        static char q[10];
        if(res==0)	putch('0');
        else if(res < 0){putch('-');res = -res;}
        int top = 0;
        while(res){
            q[top++] = res % 10 + '0';
            res /= 10;
        }
        while(top--)	putch(q[top]);
    }
    inline void space(bool x){
    	if(!x) putch('\n');
    	else putch(' ');
    }
}
inline void read(int &x){
    int rt = FastIO::getint();
    x = rt;
}
inline void print(int x,bool enter){
    FastIO::putint(x);
    FastIO::flush();
    FastIO::space(enter);
}
/* definitions */
#define type_T long long
type_T k,m,ans;
/* functions */
inline type_T quick_multi(type_T a,type_T b,type_T p){
    type_T l = a * (b >> 25ll) % p * (1ll << 25) % p;
    type_T r = a * (b & ((1ll << 25) - 1)) % p;
    return (l + r) % p;
}
inline type_T quick_power(type_T a,type_T b,type_T p){
    type_T res = 1, base = a;
    while(b){
        if(b & 1) res = quick_multi(res,base,p) % p;
        b >>= 1; base = quick_multi(base,base,p) % p;
    }
    return res;
}
inline type_T BSGS(type_T a,type_T b,type_T p){
    std::map<type_T,type_T> hash;
    hash.clear(); b %= p;
    type_T t = (type_T)std::sqrt(p) + 1;
    FORL(i,0,t-1,1){
        type_T value = (type_T)quick_multi(b,quick_power(a,i,p),p);
        hash[value] = i;
    }
    a = quick_power(a,t,p);
    if(!a){if(!b) return 1ll; else return -1;}
    FORL(i,0,t,1){
        type_T value = quick_power(a,i,p);
        type_T tmp = hash.find(value) == hash.end() ? -1 : hash[value];
        if(tmp >= 0 && i * t - tmp >= 0) return i * t - tmp;
    }
    return -1;
}
int main(int argc,char *argv[]){
    #define nine(x) (x << 3) + x
    scanf("%lld%lld",&k,&m);
    ans = BSGS(10,nine(k) + 1,m);
    printf("%lld\n",ans);
    return 0;
}

THE END