poj 3619 Speed Reading

Speed Reading

Description

All K (1 ≤ K ≤ 1,000) of the cows are participating in Farmer John‘s annual reading contest. The competition consists of reading a single book with N (1 ≤ N ≤ 100,000) pages as fast as possible while understanding it.

Cow i has a reading speed S_i (1 ≤ S_i ≤ 100) pages per minute, a maximum consecutive reading time T_i (1 ≤ T_i ≤ 100) minutes, and a minimum rest time R_i (1
≤ R_i ≤ 100) minutes. The cow can read at a rate ofS_i pages per minute, but only for T_i minutes at a time. After she stops reading to rest, she must rest for R_i minutes before commencing
reading again.

Determine the number of minutes (rounded up to the nearest full minute) that it will take for each cow to read the book.

Input

* Line 1: Two space-separated integers: N and K

* Lines 2..K+1: Line i+1 contains three space-separated integers: S_i , T_i , and R_i

Output

* Lines 1..K: Line i should indicate how many minutes (rounded up to the nearest full minute) are required for cow i to read the whole book.

Sample Input

10 3
2 4 1
6 1 5
3 3 3

Sample Output

6
7
7

题意：牛读一个厚度为n的书，每秒可以读s页，可以连续读t秒，不过读完t秒后必需要休息r秒的时间，求牛读完这本书所需的时间；

#include <iostream>
using namespace std;
int main(){
	int n,k,s,t,r;
	cin>>n>>k;
	for (int i=0;i<k;i++){
		cin>>s>>t>>r;
		int page=n;
		int time=0;
		while (page>s*t){
			time=time+t+r;
			page-=s*t;
		}
		time+=page/s;
		if (page%s)
			time++;
		cout<<time<<endl;
	}
	return 0;
}