NBUT 1449 Annie

• [1449] Annie

• 时间限制: 1000 ms　内存限制: 65535 K
• 问题描述
• Annie is a kawaii loli in League of Legends. High Councilor Kiersta Mandrake said that Annie may be one of the most powerful champions ever to have fought in a Field of Justice.

  Annie has an ability, Molten Shield(熔岩护盾). This shield increases Annie's armor and magic resist and damages enemies who hit Annie with basic attacks.

  This ability keeps for 5 seconds and cost 10 seconds to cold down to cast nextMolten Shield.

  We assume that Annie casts the Molten at the very beginning of 0 second. And she casts the next Molten Shield when it colds down immediately. The magic damage of the base attack when she is under the shield is 60 + (0.2 * Spell Power).

  We give you the spell power of Annie's and the time of enemies' each base attack. You should tell me how much damage Anniemade via the Molten Shield.

• 输入
• This problem contains several cases.
  The first line of each case contains two integers N and S, indicate the number of attack times and the spell power of Annie's. (0 ＜ N ≤ 1000, 0 ＜ s ≤ 1000)
  Then N lines followed. Each line is a decimal number that indicates the time moment of that base attack. The decimal number will not exceeds 1000.
• 输出
• You should output the magic damage via the Molten Shield. 1 decimal places reserved.
• 样例输入

• 5 301
  2.00
  5.01
  5.00
  8.00
  14.03

• 样例输出

• 360.6

• 提示

• 无

• 来源

• XadillaX

  
  
  题目意思是说，黑暗之女-安妮的E技能是一个可以造成一个魔法伤害的盾，开了盾之后就能给攻击者一个魔法伤害，盾能持续5秒，需要冷却10秒才能重新开盾，给出攻击的时间，求安妮能够造成多大的魔法伤害。
  注意：盾从开的那一刻开始冷却，与持续时间不冲突，这里可能造成误区。题目意思是，每10秒钟开一次盾，不管有没有攻击者。。。
  
  

  #include<cstdio>
  #include<cstring>
  const int mx = 1000+10;
  
  double t[mx];
  int n,S;
  int main()
  {
      while(~scanf("%d%d",&n,&S))
      {
          int cnt=0;
          for(int i=0;i<n;i++)
          {
              scanf("%lf",&t[i]);
              int a=(int)t[i];
              a%=10;
              if(a==5 && (int)t[i]<t[i]) 
                  continue;    
              //（int）有可能是向下取整，如果是向下取整刚好等于5这种情况要排除
              if(a>=0 && a<=5)    
                  cnt++;      //在盾的持续时间内攻击，累加器 ++ 
          }
          printf("%.1lf\n",(60+(S*0.2))*cnt);
      }
      return 0;
  }