2019牛客暑期多校训练营（第三场） D-Big Integer

2019牛客暑期多校训练营（第三场） D-Big Integer

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

For little pupils, a very large number usually means an integer with many many digits. Let’s define a class of big integers which consists only of the digit one(11⋯1). The first few integers in this class are 1,11,111,1111⋯. Denote A(n) as the n-th smallest integer in this class. To make it even larger, we consider integers in the form of A(a^ b). Now, given a prime number p, how many pairs (i,j) are there such that 1 ≤ i ≤ n,1 ≤ j ≤m, A(i^j) ≡0(mod p)

输入描述:

The input contains multiple cases. The first line of the input contains a single integer T (1≤T≤100), the number of cases.
For each case, the input consists of a single line, which contains 3 positive integers p, n, m (p, n, m ≤ 10 e 9) .

输出描述:

Print the answer, a single integer, in one separate line for each case.

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题意

1 ≤ i ≤ n, 1 ≤ j ≤ m 有多少种情况满足i ^ j个1能被p整除。

思路

原式：A(i ^ j) % p==0(记i ^ j为n)
易知，A(n)等价于(10 ^ n-1)/9，即为n个1。可化简为：((10^n-1)/9)%p ==0。
因为p为质数，所以9与p互质，故((10^n-1)*inv9)%p ==0。
因为inv9!=0，故(10^n-1)%p ==0，即(10 ^ n)≡1(mod p)。
根据费马小定理，10与p互质，故10 ^ phi( p)≡1(mod p)。
因为p为素数，phi(p )==p-1，所以10 ^ (p-1)≡1(mod p)。
所以最小循环节d必定是p-1的因数。然后直接O(sqrt(p-1))暴力找到d。
(最后一点懒得写了，直接贴牛客的题解了)

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代码

#include 
#define min(x,y) (x>y?y:x)
#define __bug(x) cout<<"---test:"<>=1;
    }
    return ret%p;
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll p,n,m;
        scanf("%lld%lld%lld",&p,&n,&m);
        if(p==2||p==5)
        {
            printf("0\n");
            continue;
        }
        if(p==3)
        {
            printf("%lld\n",n/3*m);
            continue;
        }
        ll d=p-1;
        for(int i=2;i*i<=p-1;i++)
        {
            if((p-1)%i==0)
            {
                if(quick_pow(10,i,p)==1)
                {
                    d=min(d,i);
                }
                if(quick_pow(10,(p-1)/i,p)==1)
                {
                    d=min(d,(p-1)/i);
                }
            }
        }
//        __bug(d);
        int cnt=0;
        ll dd=d;
        ll maxn=0;
        for(int i=2;i*i<=d;i++)
        {
            if(dd%i==0)
            {
                pr[cnt].num=i;
                pr[cnt].times=0;
                while(dd%i==0)
                {
                    dd/=i;
                    pr[cnt].times++;
                }
                maxn=max(maxn,pr[cnt].times);
                cnt++;
//                __bug(maxn);
            }
        }
        ll g=1;
        if(dd>1)
        {
            pr[cnt].num=dd;
            pr[cnt].times=1;
            maxn=max(pr[cnt].times,maxn);
            cnt++;
//            __bug(maxn);
        }
//        for(int i=0;imaxn)
        {
            ans+=n/g*(m-maxn);
        }
        printf("%lld\n",ans);
    }
    return 0;
}