HDU 4291 A Short problem 矩阵，多重函数求循环节

题意：

　According to a research, VIM users tend to have shorter fingers, compared with Emacs users.
　　Hence they prefer problems short, too. Here is a short one:
　　Given n (1 <= n <= 10¹⁸), You should solve for 

g(g(g(n))) mod 10⁹ + 7
　　where
g(n) = 3g(n - 1) + g(n - 2)
g(1) = 1
g(0) = 0

题解：

假如最外层要求模10^9+7，则令L0=10^9+7

将最外层内部看做一个整体x，则x在L0下可以求出一个循环节L1

那么x的值对L1取模不影响结果

将次外层内部看做一个整体y，则y在L1下可以求出一个循环节L2

那么y的值对L2取模不影响结果

......

求循环节的代码：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;#define lint __int64const lint mod = 1000000007;lint L1 = 222222224, L2 = 183120, L3;void test(){    printf("start\n");    lint gn, gn1 = 1, gn2 = 0;/*    for(lint i = 2; ; i++)    {        gn = (3 * gn1 + gn2) % mod;        if(gn == 1 && gn1 == 0)        {            printf("循环节L1：%I64d\n", i - 1);            L1 = i - 1; //222222224            break;        }        gn2 = gn1;        gn1 = gn;    }    gn = 3, gn1 = 1, gn2 = 0;    for(lint i = 2; ; i++)    {        gn = (3 * gn1 + gn2) % L1;        if(gn == 1 && gn1 == 0)        {            printf("循环节L2：%I64d\n", i - 1);            L2 = i - 1; //183120            break;        }        gn2 = gn1;        gn1 = gn;    }*/    gn = 3, gn1 = 1, gn2 = 0;    for(lint i = 2; ; i++)    {        gn = (3 * gn1 + gn2) % L2;        if(gn == 1 && gn1 == 0)        {            printf("循环节L3：%I64d\n", i - 1);            L3 = i - 1; //183120            break;        }        gn2 = gn1;        gn1 = gn;    }    printf("over\n");}int main(){    test();}

题解：

#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;#define MAXN 3#define lint __int64const lint L0 = 1000000007;const lint L1 = 222222224;const lint L2  = 183120;const lint L3 = 240;lint gn, mod;class CMatrix{public:    lint a[MAXN][MAXN], n;    void init(int n, int flag);    CMatrix operator+(CMatrix);    CMatrix operator*(CMatrix);    CMatrix power(lint);};void CMatrix::init(int n, int flag){    this->n = n;    memset(this->a, 0, sizeof(this->a));    if(flag == 0) return;    for(int i = 0; i < n; i++)        this->a[i][i] = 1;}CMatrix CMatrix::operator+(CMatrix tt){    CMatrix ret = *this;    for(int i = 0; i < n; i++)        for(int j = 0; j < n; j++)            ret.a[i][j] = (ret.a[i][j] + tt.a[i][j]) % mod;    return ret;}CMatrix CMatrix::operator*(CMatrix tt){    CMatrix ret;    ret.init(this->n, 0);    for(int i = 0; i < n; i++)    {        for(int j = 0; j < n; j++)        {            for(int k = 0; k < n; k++)                ret.a[i][j] += this->a[i][k] * tt.a[k][j];            ret.a[i][j] %= mod;        }    }    return ret;}CMatrix CMatrix::power(lint e){    CMatrix ret, tmp;    ret.init(this->n, 1);    tmp = *this;    while(e > 0)    {        if(e & 1)            ret = ret * tmp;        e >>= 1;        tmp = tmp * tmp;    }    return ret;}int main(){    while(scanf("%I64d", &gn) != EOF)    {        CMatrix x, y;        x.n = 2;        x.a[0][0] = 3; x.a[0][1] = 1;        x.a[1][0] = 1; x.a[1][1] = 0;        mod = L2;        y = x.power(gn % L3);        mod = L1;        y = x.power(y.a[1][0] % L2);        mod = L0;        y = x.power(y.a[1][0] % L1);        printf("%I64d\n", y.a[1][0]);    }    return 0;}