CodeForces 1883A Morning

合集 - Codeforces(21)

1.CodeForces 1992E Novice's Mistake2024-07-162.CodeForces 1935A Entertainment in MAC2024-06-163.CodeForces 1992C Gorilla and Permutation2024-07-164.CodeForces 1992D Test of Love2024-07-165.CodeForces 1992B Angry Monk2024-07-166.CodeForces 1992A Only Pluses2024-07-167.CodeForces 1983B Corner Twist2024-07-168.CodeForces 1983A Array Divisibility2024-07-169.CodeForces 1983C Have Your Cake and Eat It Too2024-07-1610.CodeForces 1883B Chemistry2024-07-26

11.CodeForces 1883A Morning2024-07-26

12.CodeForces 1883C Raspberries2024-07-2813.CodeForces 1883D In Love2024-07-2814.CodeForces 1883E Look Back2024-07-2815.CodeForces 1883F You Are So Beautiful2024-07-2816.CodeForces 1883G1 Dances (Easy version)2024-07-2817.CodeForces 908B New Year and Buggy Bot2024-08-0118.CodeForces 908C New Year and Curling2024-08-0119.CodeForces 1132B Discounts2024-08-0120.CodeForces 1619D New Year's Problem2024-08-0121.Codeforces Round 964 (Div. 4)2024-08-07

收起

题目链接：CodeForces 1883A【Morning】

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思路

       模拟，特判当密码中的某个元素为0时，用10减去当前光标的位置，并修改光标的位置为当前元素，再操作依次显示当前元素。对于其他情况则直接使用光标的位置减去目标位置，修改光标位置为当前元素，然后再操作一次显示当前元素。

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代码

#include <bits/stdc++.h>
using namespace std;
 
void solve() {
  string s;
  cin >> s;
  int now = 1, res = 0;
  for (int i = 0; i < 4; i++) {
    if (s[i] == '0')
      res += abs(now - 10) + 1, now = 10;
    else
      res += abs(now - (s[i] - '0')) + 1, now = s[i] - '0';
  }
  cout << res << endl;
}
 
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}