10692

题目链接

10692 - Huge Mods

分析

其实这个问题很简单,有以下定理可用

• 若(a,n)=1,ax≡ax mod ϕ(m)(mod m)

• 若(a,n)≠1,ax≡ax mod ϕ(m)+ϕ(m)(mod ϕ(m))

因此我们可以递归的求解这个问题.

f(aa2a3…1)≡af(a2a3…)modϕ(m)+ϕ(m)1(mod m)

注意一定要加上ϕ(m)因为当整除的时候会出现问题.

#include <cstdio>#include <iostream>#include <vector>#include <queue>#include <algorithm>#include<cmath>#include <cstring>#include <map>#include <set>#include <iomanip>#include <bitset>#define pb push_back#define mp make_pair#define fi first#define se second#define INF 0x3f3f3f3f#define INF64 0x3f3f3f3f3f3f3f3fusing namespace std;const int mod = 1e8+7;const int MAX_P = 2e4+10;const int maxn =1e5+10;const int MAX_V = 5e5+10;const int maxv = 1e6+10;typedef long long LL;typedef long double DB;typedef pair<int,int> Pair;int n;int a[maxn];int phi[maxn],prime[maxn],cnt;void phi_table() {    memset(prime,0,sizeof(prime));    cnt =0;    phi[1] = 1;    for(int i=2 ; i<maxn ; ++i)    {        if(!prime[i]){            prime[cnt++] = i;            phi[i] = i-1;        }        for(int j =0 ; j< cnt && i*prime[j] < maxn ; ++j){            prime[i*prime[j]] = 1;            if(i % prime[j])phi[i*prime[j]] = phi[i]*(prime[j]-1);            else{                phi[i*prime[j]] = phi[i]*prime[j];                break;            }        }    }}LL power_mod(LL a, LL b, LL mod) {     LL ans = 1;    while(b > 0) {        if(b & 1) ans = ans * a % mod;        a = a * a % mod;        b >>= 1;    }    return ans + mod;}LL f(int idx,LL mod){    if(idx == n-1) return a[idx] < mod ? a[idx]  : a[idx]% mod + mod;    return power_mod(a[idx],f(idx+1,phi[mod]),mod);}int main() {    // ios::sync_with_stdio(false);    // cin.tie(nuLLptr);    // cout.precision(10);    // cout << fixed;#ifdef LOCAL_DEFINE    freopen("in.txt", "r", stdin);#endif    phi_table();    // for(int i=1 ; i< 100 ; ++i)    //     std::cout << phi[i] << '\n';    int m;    int kase =0;    while (cin>>m && m != (int)'#') {        //std::cout << m << '\n';        cin>>n;        for(int i=0 ; i<n ; ++i)scanf("%d",&a[i] );        printf("Case #%d: %lld\n",++kase,f(0,m)%m );    }#ifdef LOCAL_DEFINE    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";#endif    return 0;}