在最多花费 k 的情况下，最大可能的平衡二叉子串分割数

#include <stdio.h>
#include <limits.h>
#include <string.h>
//Define a two-dimensional matrix m
int m[1001][1001];

//Define a function to find maximum possible //balanced binary substring splits
int maxSplits(int a[], int v[], int k, int s) {
   if (k < 0) {
      return INT_MIN;
   }
   if (m[k][s] != -1) {
      return m[k][s];
   }
   
   //Define integer variables to count the number of zeros and ones 
   // Define an integer variable to count the //number of splits
   int zeroCount = 0, oneCount = 0;
   int cntSplits = 0;
   int i;
   
   //Iterating through the given array a
   for (i = s - 1; i > 0; i--) {
      a[i] == 0 ? zeroCount++ : oneCount++;
      
   // check whether the number of zeros is equal to the number of ones, then store the maximum feasible one
      if (zeroCount == oneCount) {
         cntSplits = cntSplits > (1 + maxSplits(a, v, k - (v[i] - v[i - 1]) * (v[i] - v[i - 1]), i)) ? cntSplits : (1 + maxSplits(a, v, k - (v[i] - v[i - 1]) * (v[i] - v[i - 1]), i));
      }
   }
   
   //Suppose the index is at the position 0, then find whether it is a one or a zero. then increment the count
   if (i == 0) {
      a[0] == 0 ? zeroCount++ : oneCount++;
      
   // set the cntSplits to zero , if count of one and count of zero is unequal.
      if (zeroCount != oneCount) {
         cntSplits = 0;
      }
   }
   
   // store the resultant value in the matrix
   return m[k][s] = cntSplits;
}
int main() {
   int a[] = { 1, 0, 0, 1, 0, 0, 1, 1 };
   int v[] = { 7, 8, 9, 10, 11, 12, 13, 14 };
   int k = 1;
   int s = sizeof(a) / sizeof(a[0]);
   
   //To assign a specific value to a block of memory, we use the memset() function.
   memset(m, -1, sizeof(m));
   printf("%d\n", maxSplits(a, v, k, s));
   return 0;
}