HDU 3221 Brute-force Algorithm(指数降幂公式)

题目链接；
HDU 3221 Brute-force Algorithm
题意：
根据递归可以得到f[1]=a,f[2]=b,f[n]=f[n−1]∗f[n−2](n≥3),给出a,b,p,n求f[n]%p的值。
数据范围：1≤n≤1000000000,1≤P≤1000000,0≤a,b<1000000.
分析：
设菲波那切数列第n项为g[n],g[1]=0,g[2]=1,g[3]=1.根据递推可以得到:
f[n]=ag(n−1)∗bg(n) %p  (n≥3)
但是考虑到指数g(n)会很大，我们需要降幂。
指数降幂公式:

Ax % p=Ax%ϕ(p)+ϕ(p) % p （x≥ϕ(p),ϕ(p)是p的欧拉函数值）

所以用矩阵快速幂求菲波那切数列时的取模数是ϕ(p)而不是p。

下面的代码用G++提交能AC，用C++就不能。。。

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <climits>#include <cmath>#include <ctime>#include <cassert>#include <bitset>#define IOS ios_base::sync_with_stdio(0); cin.tie(0);using namespace std;typedef long long ll;const int SIZE = 10;const int MAX_N = 1000010;bitset<MAX_N> bs;int prime_cnt, prime[MAX_N], phi[MAX_N];struct Matrix {    int row, col;    ll data[SIZE][SIZE];    Matrix () {}    Matrix (int _row, int _col): row(_row), col(_col) {}};void GetPhi(){    bs.set();    prime_cnt = 0;    phi[1] = 1;    for(int i = 2; i < MAX_N; ++i) {        if(bs[i] == 1) {            prime[prime_cnt++] = i;            phi[i] = i - 1;        }        for(int j = 0; j < prime_cnt && i * prime[j] < MAX_N; ++j) {            bs[i * prime[j]] = 0;            if(i % prime[j]) {                phi[i * prime[j]] = (prime[j] - 1) * phi[i];            }else {                phi[i * prime[j]] = prime[j] * phi[i];                break;            }        }    }}Matrix matrix_mul(Matrix a, Matrix b, ll p){    Matrix res;    memset(res.data, 0, sizeof(res.data));    res.row = a.row, res.col = b.col;    for(int i = 1; i <= res.row; ++i) {        for(int j = 1; j <= res.col; ++j) {            for(int k = 1; k <= a.col; ++k) {                res.data[i][j] += a.data[i][k] * b.data[k][j];                if(res.data[i][j] > p)                     res.data[i][j] = (res.data[i][j] % p + p);            }        }    }    return res;}Matrix matrix_quick_pow(Matrix a, ll b, ll p){    Matrix res, tmp = a;    memset(res.data, 0, sizeof(res));    res.row = res.col = a.row;    for(int i = 1; i <= res.row; ++i) { res.data[i][i] = 1; }    while(b) {        if(b & 1) res = matrix_mul(res, tmp, p);        tmp = matrix_mul(tmp, tmp, p);        b >>= 1;    }    return res;}ll quick_pow(ll a, ll b, ll c){    ll res = 1, tmp = a % c;    while(b) {        if(b & 1) res = res * tmp % c;        tmp = tmp * tmp % c;        b >>= 1;    }    return res;}int main(){    GetPhi();    int T, cases = 0;    ll a, b, p, n;    scanf("%d", &T);     while(T--) {        scanf("%lld%lld%lld%lld", &a, &b, &p, &n);        if(n == 1) {            printf("Case #%d: %lld\n", ++cases, a % p);            continue;        } else if(n == 2) {            printf("Case #%d: %lld\n", ++cases, b % p);            continue;        } else if(n == 3) {            printf("Case #%d: %lld\n", ++cases, a * b % p);            continue;        } else if(a == 0 || b == 0 || p == 1) {            printf("Case #%d: %d\n", ++cases, 0);            continue;        }        Matrix res, tmp;        tmp.row = tmp.col = 2;        tmp.data[1][1] = 0, tmp.data[1][2] = 1;        tmp.data[2][1] = 1, tmp.data[2][2] = 1;        res.row = 2, res.col = 1;        res.data[1][1] = 0, res.data[2][1] = 1;        tmp = matrix_quick_pow(tmp, n - 2, phi[p]);        res = matrix_mul(tmp, res, phi[p]);        ll ans = quick_pow(a, res.data[1][1], p);        ans = ans * quick_pow(b, res.data[2][1], p) % p;        printf("Case #%d: %lld\n", ++cases, ans);    }    return 0;}