- 守望者的逃离
- @ 2015-08-20 09:27:52
//Escape  Greedy
    #include<iostream>
    using namespace std;
int m,s,t,s1,t1;
    int main(void)
    {
        cin>>m>>s>>t;
        t1=t; 
        s1=0; 
        while (m>=10 && s-s1>0 && t1>0) 
        {
            m-=10;
            s1+=60;
            t1--;
        } 
        while (m>=6 && s-s1>17 && t1>1)
        {
            m+=4-10;
            s1+=60;
            t1-=2;
        }
        while (m>=2 && s-s1>34 && t1>2) 
        {
            m+=8-10;
            s1+=60;
            t1-=3; 
        }
        while (s-s1>119 && t1>=7)
        {
            s1+=120;t1-=7;
        }
        while (t1>0 && s-s1>0) {t1--;s1+=17;} 
      if (s1>=s) { cout<<"Yes"<<endl; cout<<t-t1<<endl;}
               else cout<<"No"<<endl<<s1<<endl;
        return 0;
    }
1 条评论
- 
  niyinchen LV 9 @ 2015-08-24 16:02:59弱弱的问一句:本题正解不是 动态规划 吗 
- 1