UVa 10692 Huge Mods

方法：数论

其实不是很明白，为什么这个公式可行

a^b % m = a^(b%phi[m] + phi[m]) % m

code:

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <string>
#include <vector>
#include <stack>
#include <bitset>
#include <cstdlib>
#include <cmath>
#include <set>
#include <list>
#include <deque>
#include <map>
#include <queue>
#include <fstream>
#include <cassert>
#include <unordered_map>
#include <unordered_set>
#include <cmath>
#include <sstream>
#include <time.h>
#include <complex>
#include <iomanip>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define FOR(a,b,c) for (ll (a)=(b);(a)<(c);++(a))
#define FORN(a,b,c) for (ll (a)=(b);(a)<=(c);++(a))
#define DFOR(a,b,c) for (ll (a)=(b);(a)>=(c);--(a))
#define FORSQ(a,b,c) for (ll (a)=(b);(a)*(a)<=(c);++(a))
#define FORC(a,b,c) for (char (a)=(b);(a)<=(c);++(a))
#define FOREACH(a,b) for (auto &(a) : (b))
#define rep(i,n) FOR(i,0,n)
#define repn(i,n) FORN(i,1,n)
#define drep(i,n) DFOR(i,n-1,0)
#define drepn(i,n) DFOR(i,n,1)
#define MAX(a,b) a = Max(a,b)
#define MIN(a,b) a = Min(a,b)
#define SQR(x) ((LL)(x) * (x))
#define Reset(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define all(v) v.begin(),v.end()
#define ALLA(arr,sz) arr,arr+sz
#define SIZE(v) (int)v.size()
#define SORT(v) sort(all(v))
#define REVERSE(v) reverse(ALL(v))
#define SORTA(arr,sz) sort(ALLA(arr,sz))
#define REVERSEA(arr,sz) reverse(ALLA(arr,sz))
#define PERMUTE next_permutation
#define TC(t) while(t--)
#define forever for(;;)
#define PINF 1000000000000
#define newline '
'

#define test if(1)if(0)cerr
using namespace std;
using namespace std;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int,int> ii;
typedef pair<double,double> dd;
typedef pair<char,char> cc;
typedef vector<ii> vii;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> l4;
const double pi = acos(-1.0);

int phi(int n)
{
    int ret = n;
    for (int i = 2; i * i <= n; ++i)
    {
        if (n % i) continue;
        ret -= ret/i;
        while (n % i == 0) n /= i;
    }
    if (n != 1)
        ret -= ret/n;
    return ret;
}
int m, n;
int a[10];
ll pow_mod(ll base, ll p, ll mod)
{
    if (!base) return 0;
    ll ret = 1;
    while (p)
    {
        if (p & 1) ret = ret * base % mod;
        p >>= 1;
        base = base * base % mod;
    }
    return ret;
}
int dfs(int cur, int mod)
{
    if (cur == n-1) return a[cur] % mod;
    int ph = phi(mod);
    return pow_mod(a[cur], dfs(cur+1, ph)%ph + ph, mod);
}
int main()
{
    int kase = 0;
    while (cin >> m >> n)
    {
        rep(i, n) cin >> a[i];
        cout << "Case #" << ++kase << ": " << dfs(0, m) << newline;
    }
}
/*
 10 4 2 3 4 5
 100 2 5 2
 53 3 2 3 2
 #
*/