圈乘运算问题

#include "iostream"#include "algorithm"#include "cmath"#include "fstream"using namespace std;#define INF 100000/*求X和圈乘运算组成的值为K的表达式最少需要多少个圈乘运算  m为X的位数，圈乘运算表达式最大值不超过 L = (9*m)*9 + 9，当m=1时，不超过171  m = lg(X+1) 向上取整  s[L][4]存储各数字信息  s[i][0] 存储i所需最少圈乘运算次数，s[i][1]存储sum(i)，s[i][2]存储max(i)，s[i][3]存储min(i)  s[0][0]、s[0][1]、s[0][2]、s[0][3]分别存储X信息    s[K][0]为结果*/int **s;int X, K, L;void count(int num[], int i){    int t;    int max = 0;    int min = 10;    int sum = 0;    while(i)    {        t = i % 10;        if(t > max)            max = t;          if(t < min)            min = t;          sum += t;             i /= 10;    }    num[1] = sum;    num[2] = max;    num[3] = min;}void init(){    int m = ceil(log(X+1)/log(10.0));     L = 81 * m + 9;    if(L < 171)        L = 171;    s = new int*[L+1];    for(int i=0; i<=L; i++)        s[i] = new int[4];    for(i=1; i<=L; i++)    {        s[i][0] = INF;  //圈乘运算最少次数        s[i][1] = 0;    //sum(i)        s[i][2] = 0;    //max(i)        s[i][3] = INF;  //min(i)    }    for(i=1; i<=L; i++)  //计算s[i][4]中的信息        count(s[i], i);     s[0][0] = 0;    count(s[0], X);}int dyna(){    init();    if(X == K)        return 0;    if(K > L)        return -1;    bool flag = true;     while(flag)  //最优值不再更新时退出    {        flag = false;        for(int i=0; i<=L; i++)        {            if(s[i][0] < INF)  //i已计算过                for(int j=0; j<=L; j++)                    if(s[j][0] < INF)  //j已计算过                    {                        int a = s[i][1] * s[j][2] + s[j][3];                        if(s[a][0] > s[i][0] + s[j][0] + 1) //圈乘个数更少，更新                        {                            s[a][0] = s[i][0] + s[j][0] + 1;                             flag = true;                         }                    }        }    }    if(s[K][0] < INF)        return s[K][0];    else        return -1;}int main(){    ifstream fin("圈乘.txt");    cout << "输入整数X：";    fin >> X; cout << X;    cout << "\n输入整数K：";    fin >> K; cout << K;    cout << "\n最少圈乘运算个数为：" << dyna() << endl;    fin.close();    return 0;}