数値を 0 にするために必要な最小演算数を求める C++ プログラム

n := size of S
x := digit at place S[n - 1]
for initialize i := 0, when i <= n - 2, update (increase i by 1), do:
   if S[i] is not equal to &#39;0&#39;, then:
      x := x + (digit at place S[i]) + 1
return x

#include <bits/stdc++.h>
using namespace std;

int solve(string S) {
   int n = S.size();
   int x = S[n - 1] - &#39;0&#39;;
   for (int i = 0; i <= n - 2; i++)
      if (S[i] != &#39;0&#39;)
         x = x + S[i] + 1 - &#39;0&#39;;
   return x;
}
int main() {
   string S = "1000";
   cout << solve(S) << endl;
}

"1000"

2