poj 3619 Speed Reading

Speed Reading

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 8431 Accepted: 3935

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 32 4 16 1 53 3 3

Sample Output

677

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;}