poj2720 Last Digits

这题.....简直没话可说
首先需要利用欧拉定理优化指数 防止快速幂爆精度
然后需要注意 每次使用的欧拉函数大小都不一样
最后输出的时候补0....
然后特判 指数小于phi(m)的数不能约 要单独算(或者叫打表?)
Code:

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = 300000;

ll b,n,p,tot,num[10],cnt,ans,f[110][110];
int prime[1000010],phi[10000010];
bool vis[10000010];

ll qpow(ll a,ll b,ll mod)
{
    ll res = 1;
    while(b)
    {
        if (b & 1)
            res = (res * a) % mod;
        b >>= 1;
        a = (a * a) % mod;
    }
    return res;
}

void init()
{
    phi[1] = 1;
    phi[10000000] = 4000000;
    phi[4000000] = 1600000;
    phi[1600000] = 640000;
    phi[640000] = 256000;
    for (int i = 2; i <= maxn; i++)
    {
        if (!vis[i])
        {
            prime[++tot] = i;
            phi[i] = i - 1;
        }
        for (int j = 1; j <= tot; j++)
        {
            ll t = prime[j] * i;
            if (t > maxn)
                break;
            vis[t] = 1;
            if (i % prime[j] == 0)
            {
                phi[t] = phi[i] * prime[j];
                break;
            }
            phi[t] = phi[i] * (prime[j] - 1);
        }
    }
}

ll solve(ll dep,ll mod)
{
    if (dep == 1)
        return b;
    if (mod == 1)
        return 0;
    if (f[b][dep] == -1)
    {
        ll temp = phi[mod];
        ll tt,temp2 = solve(dep - 1,temp);
        tt = qpow(b,temp2 + temp,mod);
        return f[b][dep] = tt;
    }
    return f[b][dep] % mod;
}

int main()
{
    memset(f,-1,sizeof(f));
    init();
    while (~scanf("%lld",&b) && b)
    {
        cnt = 0;
        memset(num,0,sizeof(num));
        scanf("%lld%lld",&n,&p);
        if (p == 0)
            printf("0
");
        else
        {
            ll cur = p;
            p = qpow(10,p,100000010);
            if (f[b][n] != -1)
                ans = f[b][n];
            else
            {
                if (n == 2 && b <= 6)
                    ans = qpow(b,b,10000000);
                else
                if (n == 3 && b <= 3)
                    ans = qpow(b,qpow(b,b,10000000),10000000);
                    else
                        if (n == 4 && b == 2)
                            ans = qpow(b,qpow(b,qpow(b,b,10000000),10000000),10000000);
                    else
                ans = solve(n,10000000);
                f[b][n] = ans;
            }
            while (ans)
            {
                num[++cnt] = ans % 10;
                ans /= 10;
            }
            for (int i = cur; i >= 1; i--)
                printf("%d",num[i]);
            printf("
");
        }
    }

    return 0;
}