题解P4884 多少个1

P4884 多少个1？

一道比较裸的BSGS题

根据题意得出 \(10^n+10^{n-1}+...+1\equiv k(mod~m)\Rightarrow \frac{10^n-1}{9}\equiv k(mod~m)\Rightarrow 10^n\equiv 9k+1(mod~m)\)

直接套BSGS板子就行了,注意在乘的过程中可能爆long long,加个快速乘,时间复杂度 \(O(\sqrt{n}\log{n})\)

代码

#include<iostream>
#include<cstdio>
#include<unordered_map>
#include<cmath>
using namespace std;
typedef long long ll;

ll ksc(ll a, ll b, ll p)
{
    ll res = 0;
    while (b)
    {
        if (b & 1)
            res = (res + a) % p;
        a = (a + a) % p;
        b >>= 1;
    }
    return res;
}

ll ksm(ll a, ll b, ll p)
{
    ll res = 1;
    while (b)
    {
        if (b & 1)
            res = ksc(res, a, p);
        a = ksc(a, a, p);
        b >>= 1;
    }
    return res;
}

ll BSGS(ll a, ll b, ll p)
{
    if (1 % p == b % p)
        return 0;
    unordered_map<ll, ll> hash;
    ll k = sqrt(p) + 1;
    for (ll i = 0, j = b % p; i < k; i++, j = ksc(j, a, p))
        hash[j] = i;
    ll ak = ksm(a, k, p);
    for (ll i = 1, j = ak; i <= k; i++, j = ksc(j, ak, p))
        if (hash.count(j))
            return i * k - hash[j];
    return -1;
}
    
int main()
{
    ll k, m;
    scanf("%lld%lld", &k, &m);
    printf("%lld", BSGS(10, (9 * k + 1) % m, m));
}