计蒜客 方程的解数 dfs

题目：

https://www.jisuanke.com/course/2291/182237

思路：

来自：https://blog.csdn.net/qq_29980371/article/details/76599695

dfs(int cnt, int curVal)//cnt是k,p的下标，curVal当前的和值
从1到m遍历进入dfs，然后不停dfs，进行查找，满足条件将结果加一，cnt临界就跳出循环，

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<queue>
 using namespace std;
 //计蒜客 方程的解数
 int n, m, k[], p[], sum;
 int poww[][];
 void init()
 {
     //memset(poww, 1, sizeof(poww));
     for (int i = ; i <= ; i++)
         poww[i][] = ;
     for (int i = ; i <= ; i++)
     {
         for (int j = ; j <= ; j++)
         {
             poww[i][j] = i*poww[i][j - ];
         }
     }
 }

 void dfs(int cnt, int curVal)//cnt是k,p的下标，curVal当前的和值
 {
     if (cnt == n  && curVal == )//满足方程式，sum++
     {
         sum++;
         return;
     }
     if (cnt == n )//跳出循环
         return;
     for (int i = ; i <= m; i++)
     {
         dfs(cnt + , curVal + k[cnt] * poww[i][p[cnt]]);
     }
 }

 int main()
 {
     while (scanf("%d%d", &n, &m) == )
     {
         for (int i = ; i<n; i++)
             scanf("%d%d", &k[i], &p[i]);

         sum = ;
         init();
         dfs(, );
         printf("%d\n", sum);
     }
     return ;
 }