HDU 4990 / BC 8B Reading comprehension

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题意要求输出题目中代码的输出

代码很好理解 i从1到n i为奇数时ans=ans*2+1 i为偶数时ans*=2

可以把两步操作合成到一起即 ans=ans*4+2 然后执行n/2次 (n为奇数就再补上一次即可)

然后用矩阵快速幂就可以了

#include <iostream>#include <cstring>#include <string>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <queue>#include <map>#include<set>#include<stack>#define bug puts("bugbugbug");using namespace std;typedef long long ll;const int maxn = 20;const int maxm = 20;ll mod, k;struct Matrix{    int n, m;    long long a[maxn][maxm];    void clear()    {        n = m = 0;        memset(a, 0, sizeof(a));    }    Matrix operator * (const Matrix &b) const   //实现矩阵乘法    {        Matrix tmp;        tmp.n = n;        tmp.m = b.m;        for (int i = 0; i < n; i++)            for (int j = 0; j < b.m; j++) tmp.a[i][j] = 0;        for (int i = 0; i < n; i++)            for (int j = 0; j < m; j++)            {                if (!a[i][j]) continue;                for (int k = 0; k < b.m; k++)                    tmp.a[i][k] += a[i][j] * b.a[j][k], tmp.a[i][k] %= mod;            }        return tmp;    }    void Copy(const Matrix &b)    {        n = b.n, m = b.m;        for (int i = 0; i < n; i++)            for(int j = 0; j < m; j++) a[i][j] = b.a[i][j];    }    void unit(int sz)    {        n = m = sz;        for (int i = 0; i < n; i++)        {            for (int j = 0; j < n; j++) a[i][j] = 0;            a[i][i] = 1;        }    }};Matrix A, B,C,D;void init(){    A.clear();    A.n=A.m=2;    A.a[0][0]=4;    A.a[0][1]=2;    A.a[1][0]=0;    A.a[1][1]=1;    D.clear();    D.n=D.m=2;    D.a[0][0]=2;    D.a[0][1]=1;    D.a[1][0]=0;    D.a[1][1]=1;    B.clear();    B.n = 2;    B.m = 1; //矩阵B是f(x)的前10个数    B.a[0][0] = 0;    B.a[1][0]=1;}Matrix Matrix_pow(Matrix A, ll k, ll mod)   //矩阵快速幂{    Matrix res;    res.clear();    res.n = res.m = A.n;    for (int i = 0; i < A.n; i++) res.a[i][i] = 1;    while(k)    {        if (k & 1) res.Copy(A * res);        k >>= 1;        A.Copy(A * A);    }    return res;}int main(){    long long  n,m;    init();    while(~scanf("%lld%lld",&n,&m))    {        mod=m;        int flag=0;        if(n%2)        {            n-=1;            flag++;        }        C.Copy(Matrix_pow(A,n/2,m));            C.Copy(C*B);        if(flag)        C.a[0][0]=(C.a[0][0]*2+1)%m;        printf("%d\n",C.a[0][0]);    }    return 0;}

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