#301 (Div.2) A. Combination Lock

1. Title Description: Click to open the link
2. Thinking: The problem requires the current password after the minimum rotation and the same as the final password. Can be solved by greedy method. Consider the I-bit, assuming that the current bit a, the target is B, then there are two modes of rotation, the required number of steps are ABS (A-B) and 10-abs (A, A, a), whichever is smaller. The minimum number of steps is the sum of the increments.
3. Code:

#define _crt_secure_no_warnings#include <iostream>#include <algorithm>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#include <map>#include <queue>#include <deque>#include <cstdlib>#include <cstdio>#include <cstring>#include <cmath>#include <ctime>#include <functional>using namespace STD;typedef Long Longll#define ME (s) memset ((s), 0,sizeof (s))#define FOR (i,n) for (int i=0;i< (n); i++)intMain () {//freopen ("T.txt", "R", stdin);    intN while(~scanf("%d", &n)) {stringSTR1, str2;Cin>> str1 >> str2;intsum =0; for(inti =0; I < n; i++) {intm =ABS(Str1[i]-str2[i]); m = min (M,Ten-M);        sum + = m; }cout<< sum << Endl; }return 0;}

#301 (Div.2) A. Combination Lock