「游族杯」上海市高校程序设计邀请赛暨华东师范大学第九届 ECNU Coder 程序设计竞赛

A.足球锦标赛

题目地址：http://acm.ecnu.edu.cn/contest/16/problem/A/

思路：数字增加1翻一次，经过一个两位数的整数翻10次，如从9翻到10需要翻10次，即1+9，经过一个三位数的整数翻19次，如从99到100需要翻19次，即1+9+9

AC代码：

import java.util.Scanner;

public class Main {
  
  public static void main(String[] args) {
    
    int [] num = new int[1000010];
    Scanner in = new Scanner(System.in);
    int t = in.nextInt();
    for (int time = 1; time<=t; time++){
      String str1 = in.next();
      String str2 = in.next();
      int k = in.nextInt();
      int num1 = Integer.valueOf(str1);
      int num2 = Integer.valueOf(str2);
      int max1 = -1;
      for (int i=0; i<=k; i++) {
        int x1 = 0;
        int x2 = 0;
        for (int j=num1+1; j<=num1+i; j++) {
          if(j % 10 == 0)
            x1++;
          if(j % 100 == 0)
            x2++;
        }
        int sum1 = i + x1 * 9 + x2 * 9;
        int y1 = 0;
        int y2 = 0;
        for (int j=num2+1; j<=num2+k-i; j++) {
          if(j % 10 == 0)
            y1++;
          if(j % 100 == 0)
            y2++;
        }
        int sum2 = k - i + y1 * 9 + y2 * 9;
        if(sum1 + sum2 > max1)
          max1 = sum1 + sum2;
      }
      System.out.println("Case " + time + ": " + max1);
    }
  }

}