POJ3619 Speed Reading【水题】

Speed Reading
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 9878 Accepted: 4654

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 of Si 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

Source

USACO 2007 November Bronze

问题链接： POJ3619 Speed Reading
问题简述：给定2个正整数N和K，分别表示书的页数和测试数据个数。每组测试数据包含S、T和R3个数据，分别表示读书速度、连续读书时间和中间歇息时间，求读完书的时间？
问题分析：简单的计算问题，不解释。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C++语言程序如下：

/* POJ3619 Speed Reading */#include <cstdio>
#include <cmath>int main()
{int n, k, s, t, r;scanf("%d%d", &n, &k);for(int i = 1; i <= k; i++) {scanf("%d%d%d", &s, &t, &r);int tot = ceil((double)n / s);printf("%d\n", (tot - 1) / t * r + tot);}return 0;
}

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