Problem 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 Si (1 ≤ Si ≤ 100) pages per minute, a maximum consecutive reading time Ti (1 ≤ Ti ≤ 100) minutes, and a minimum rest time Ri (1 ≤ Ri ≤ 100) minutes. The cow can read at a rate ofSi pages per minute, but only for Ti minutes at a time. After she stops reading to rest, she must rest for Ri 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: Si , Ti , and Ri

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

題意：一本書，頁數為n，每秒讀s頁，可連讀t秒，但讀t秒后需休息r秒，求讀完書所需時間。

思路：水題一枚，讀的懂題就行。。。

Source Program

#include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<string> #include<cstdlib> #include<queue> #include<vector> #define INF 0x3f3f3f3f #define PI acos(-1.0) #define N 100001 #define MOD 2520 #define E 1e-12 using namespace std; int main() {int n,k;scanf("%d%d",&n,&k);for(int i=0;i<k;i++){int s,t,r;scanf("%d%d%d",&s,&t,&r);int num=n;//未讀頁數int time=0;//用時while(num>s*t)//當未讀頁數>連讀t秒的頁數時{time+=t+r;//總時間=讀書時間+休息時間num-=s*t;//未讀頁數=原頁數-已讀頁數}/*處理未計算頁數*/time+=num/s;if(num%s)time++;cout<<time<<endl;}return 0; }

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