【打表找规律+矩阵快速幂】number number number HDU

Think:
1知识点：打表找规律+矩阵快速幂
2题意：给定一个数n，若可以从斐波那契数列中寻找不超过k个数(可重发选择)使得这k个数的累加和为n，则称n为mjf−good，若找不到不超过k个数(可重复选择)使得这k个数的累加和为n，则称n为mjf−bad，输入k(1≤k≤109)，输出最小的mjf_bad modulo 998244353。
3方法：
1>打表找规律，发现F[k] = f[2*(k+1)+1] - 1
2>k太大，通过矩阵快速幂求解斐波那契数列第n项

矩阵快速幂求解斐波那契数列——建议参考博客
4反思：
1>矩阵乘法注意不要爆int
2>矩阵乘法注意及时取模

vjudge题目链接

以下为Wrong Answer代码——矩阵乘法时爆int导致运算错误

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int mod = 998244353;struct Matrix{    int v[4][4];};Matrix multiply(Matrix x, Matrix y, int Matrix_len);Matrix matrix_pow(Matrix a, int k, int Matrix_len);int main(){    int k, kk, ans;    Matrix x, y, z;    while(~scanf("%d", &k)){        x.v[0][0] = 1, x.v[0][1] = 1, x.v[1][0] = 1, x.v[1][1] = 0;        y.v[0][0] = 1, y.v[0][1] = 0, y.v[1][0] = 0, y.v[1][1] = 0;        kk = 2*(k+1)+1;        x = matrix_pow(x, kk-1, 2);        z = multiply(x, y, 2);        ans = (z.v[0][0]-1+mod) % mod;        printf("%d\n", ans);    }    return 0;}Matrix multiply(Matrix x, Matrix y, int Matrix_len){    Matrix z;    memset(z.v, 0, sizeof(z.v));    for(int i = 0; i < Matrix_len; i++){        for(int j = 0; j < Matrix_len; j++){            for(int k = 0; k < Matrix_len; k++){                z.v[i][j] += x.v[i][k] * y.v[k][j];                z.v[i][j] %= mod;            }        }    }    return z;}Matrix matrix_pow(Matrix a, int k, int Matrix_len){    Matrix b;    for(int i = 0; i < Matrix_len; i++){        for(int j = 0; j < Matrix_len; j++){            i == j? b.v[i][j] = 1: b.v[i][j] = 0;        }    }    while(k){        if(k & 1)            b = multiply(b, a, 2);        a = multiply(a, a, 2);        k >>= 1;    }    return b;}

以下为Accepted代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;const int mod = 998244353;struct Matrix{    LL v[4][4];};Matrix multiply(Matrix x, Matrix y, int Matrix_len);/*矩阵乘法*/Matrix matrix_pow(Matrix a, int k, int Matrix_len);/*矩阵快速幂*/int main(){    LL k, kk, ans;    Matrix x, y, z;    while(~scanf("%lld", &k)){        x.v[0][0] = 1, x.v[0][1] = 1, x.v[1][0] = 1, x.v[1][1] = 0;        y.v[0][0] = 1, y.v[0][1] = 0, y.v[1][0] = 0, y.v[1][1] = 0;        kk = 2*(k+1)+1;        x = matrix_pow(x, kk-1, 2);        z = multiply(x, y, 2);        ans = (z.v[0][0]-1+mod) % mod;        printf("%lld\n", ans);    }    return 0;}Matrix multiply(Matrix x, Matrix y, int Matrix_len){    Matrix z;    memset(z.v, 0, sizeof(z.v));    for(int i = 0; i < Matrix_len; i++){        for(int j = 0; j < Matrix_len; j++){            for(int k = 0; k < Matrix_len; k++){                z.v[i][j] += x.v[i][k] * y.v[k][j];                z.v[i][j] %= mod;            }        }    }    return z;}Matrix matrix_pow(Matrix a, int k, int Matrix_len){    Matrix b;    for(int i = 0; i < Matrix_len; i++){        for(int j = 0; j < Matrix_len; j++){            i == j? b.v[i][j] = 1: b.v[i][j] = 0;        }    }    while(k){        if(k & 1)            b = multiply(b, a, 2);        a = multiply(a, a, 2);        k >>= 1;    }    return b;}