#1498 : Diligent Robots
Time Limit:10000ms
Case Time Limit:1000ms
Memory Limit:256MB
Description
There are N jobs to be finished. It takes a robot 1 hour to finish one job.
At the beginning you have only one robot. Luckily a robot may build more robots identical to itself. It takes a robot Q hours to build another robot.
So what is the minimum number of hours to finish N jobs?
Note two or more robots working on the same job or building the same robot won't accelerate the progress.
Input
The first line contains 2 integers, N and Q.
For 70% of the data, 1 <= N <= 1000000
For 100% of the data, 1 <= N <= 1000000000000, 1 <= Q <= 1000
Output
The minimum number of hours.
- Sample Input
- 10 1
- Sample Output
- 5
题意就是机器人完成工作，机器人可以做两种事，一个是机器人可以花一小时完成一项工作，另一个是机器人花Q小时再复制一个机器人，要求出完成工作花费的最少时间。
可以先把所有要复制的先复制完，然后再完成工作。
注意两个或两个以上的机器人在同一个工作或复制相同的机器人不会加速进展。
代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 int main(){ 5 ll n,m,a,b,ans,cnt; 6 while(~scanf("%lld%lld",&n,&m)){ 7 a=2;b=1;ans=n; //最初假设复制了一次，两个机器人 8 while(a<n){ 9 if(n%a==0) //判断一下是否整除工作量，不整除就为一个小时 10 cnt=0; 11 else 12 cnt=1; 13 ans=min(ans,b*m+n/a+cnt); 14 a*=2; //几个机器人 15 b++; //复制几次 16 } 17 printf("%lld ",ans); 18 } 19 return 0; 20 }