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  • cf 1182 E

    • 当时脑残了, 不会写矩阵快速幂中更改的系数, 其实把他扔到矩阵里同时递推就好了
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<queue>
    #include<iostream>
    #define ll long long
    using namespace std;
    ll read() {
    	ll nm = 0, f = 1;
    	char c = getchar();
    	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
    	return nm * f;
    }
    const int mod = 1000000006, mod1 = 1000000007;
    void add(int &x, int y) {
    	x += y;
    	x -= x >= mod ? mod : 0;
    }
    int mul(int a, int b) {
    	return 1ll * a * b % mod;
    }
    
    void add1(int &x, int y) {
    	x += y;
    	x -= x >= mod ? mod : 0;
    }
    
    int mul1(int a, int b) {
    	return 1ll * a * b % mod1;
    }
    
    struct Note {
    	int a[5][5], h, l;
    	Note() {
    		memset(a, 0, sizeof(a));
    		h = l = 0;
    	}
    	void init() {
    		memset(a, 0, sizeof(a));
    		h = l = 0;
    	}
    } be, ed, tmp, biao;
    
    Note operator * (Note a, Note b) {
    	Note c;
    	c.h = a.h, c.l = b.l;
    	for(int i = 0; i < a.h; i++) {
    		for(int j = 0; j < a.l; j++) {
    			for(int k = 0; k < b.l; k++) {
    				add(c.a[i][k], mul(a.a[i][j], b.a[j][k]));
    			}
    		}
    	}
    	return c;
    }
    int poww(int a, int b) {
    	int ans = 1, tmp = a;
    	for(; b; b >>= 1, tmp = mul1(tmp, tmp)) if(b & 1) ans = mul1(ans, tmp);
    	return ans;
    }
    Note poww(ll x) {
    	tmp = biao;
    	for(; x; x >>= 1, tmp = tmp *  tmp) if(x & 1) be = be * tmp;
    	return be;
    }
    
    int f1, f2, f3, c, ans = 1;
    ll n;
    void init() {
    	be.init();
    	biao.init();
    }
    
    int main() {
    	n = read();
    	n -= 3;
    	f1 = read(), f2 = read(), f3 = read(), c = read();
    	init();
    	be.h = 1, be.l = 3;
    	be.a[0][0] = 1;
    	biao.h = biao.l = 3;
    	biao.a[0][2] = biao.a[1][0] = biao.a[1][2] = biao.a[2][1] = biao.a[2][2] = 1;
    	ed = poww(n);
    	ans = mul1(ans, poww(f1, ed.a[0][2]));
    	init();
    	be.h = 1, be.l = 3;
    	be.a[0][1] = 1;
    	biao.h = biao.l = 3;
    	biao.a[0][2] = biao.a[1][0] = biao.a[1][2] = biao.a[2][1] = biao.a[2][2] = 1;
    	ed = poww(n);
    	ans = mul1(ans, poww(f2, ed.a[0][2]));
    	init();
    	be.h = 1, be.l = 3;
    	be.a[0][2] = 1;
    	biao.h = biao.l = 3;
    	biao.a[0][2] = biao.a[1][0] = biao.a[1][2] = biao.a[2][1] = biao.a[2][2] = 1;
    	ed = poww(n);
    	ans = mul1(ans, poww(f3, ed.a[0][2]));
    	init();
    	be.h = 1, be.l = 5;
    	be.a[0][3] = 8, be.a[0][4] = 1;
    	biao.h = biao.l = 5;
    	biao.a[4][4] = biao.a[1][0] = biao.a[0][2] = biao.a[1][2] = biao.a[2][1] = biao.a[2][2] = biao.a[3][2] = biao.a[3][3] = 1;
    	biao.a[4][2] = mod - 6;
    	biao.a[4][3] = 2;
    	ed = poww(n);
    	ans = mul1(ans, poww(c, ed.a[0][2]));
    	cout << ans << "
    ";
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/luoyibujue/p/11010257.html
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