[UVA][循] 12464 - Professor Lazy, Ph.D.＠Morris' Blog｜PChome Online 人新台

Morris' Blog 我在摸索我存在的意、生命的意、涵及值。有真正神手般的害，套用模式是大部分的情。那害的我存在的意何，托出神的存在？有候不能怪人不我，我已始省思，什我有人的原因，而是全要靠人。有候我我不同，那是因我做不到很多人能到的事情。我不到、玩不起、得不明、理解得不多，我缺少的文能力，就是一切一切能的判定。此，我不得能做些人有助的事情。我知道我自己是不服的性格，但在我不得不。我有能力，因此我必。我到底世界作什？充不定性，而我是法出拔萃，想一般人似乎也回不了了。越大的挑需要越力的支持，而我的支持是什？最近人，但都是失的局，人能接受，不需要言，不需要通，一切所想都表不出。
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2013-06-05 10:38:47| 人527| 回0 | 上一篇 | 下一篇

[UVA][循] 12464 - Professor Lazy, Ph.D.

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Problem D. Professor Lazy, Ph.D.

Professors are very motivated when they have to travel abroad for a conference (of course, if fees are paid by the university), but they don’t have the same attitude when the moment to grade exams arrives.

Professor Lazy, Ph.D., has a particular way to grade exams (and very unfair, by the way). He puts all his exams in a box and then starts getting them out one by one in a totally random fashion. He assigns grade α to the first exam that he gets out of the box, and grade β to the second exam that he gets out of the box. From that point on, he assigns a grade to each of the exams based on the grades of the previous two exams. What he does is that he takes the grade of the immediately previous exam, adds 1 and divides by the grade of the exam before the previous one.

For example, let’s imagine that α = 2 and β = 3. This is what happens:

The first exam gets α = 2.
The second exam gets β = 3.
The previous two grades are α and β, so the third exam gets = = 2.
The previous two grades are β and , so the fourth exam gets = = 1.
The procedure continues until he’s done with all exams.

More formally, we can define the grade Q_n of the nth exam with a recurrence like this:

Q_n =

( || α if n = 0, |{ β if n = 1, |||( 1-+-Qn--1 Qn -2 if n ≥ 2.

Even this simple procedure is a lot of work for Professor Lazy, Ph.D., so he asks you to write a program to do it for him. He wants to spend all day long drinking coffee in the cafeteria with other professors. Given α, β and n find the value of Q_n.

Note that the grades do notnecessarily lie inside a fixed range. They are just arbitrary integers.

Input

The input contains several test cases (at most 1000). Each test case is described by three integer numbers α, β and n on a single line (1 ≤ α, β ≤ 10⁹ and 0 ≤ n ≤ 10¹⁵).

The last line of the input contains three zeros and should not be processed.

Output

For each test case, write the value of Q_n in a single line. The input will be such that the value of Q_n is always an integer. Furthermore, Q_i will never be zero for 0 ≤ i ≤ n (in other words, division by zero will never arise when evaluating the recurrence).

Sample input and output

standard input	standard output

1 1 0 1 2 1 5 9 2 2 3 3 7 4 4 2109 650790 344341899059516 45861686 57 594261603792314 2309734 21045930 808597262407955 0 0 0	1 2 2 1 2 650790 804591 2309734

Andrés Mejía-Posada, May 2012

有，看起超解的。
但是使用代的方式，去求 Q0, Q1, Q2, ... 就
Q5 居然 = alpha, 想到又回了。

直接知道五一循，然後打表理之。

#include <stdio.h>

int main() {
    int n, a, b;
    long long m;
    while(scanf("%d %d %lld", &a ,&b, &m) == 3) {
        if(m == 0 && a == 0 && b == 0)
            break;
        n = m%5;
        if(n == 0)
            printf("%d\n", a);
        if(n == 1)
            printf("%d\n", b);
        if(n == 2)
            printf("%d\n", (1+b)/a);
        if(n == 3)
            printf("%d\n", (a+b+1)/a/b);
        if(n == 4)
            printf("%d\n", (a+1)/b);
    }
    return 0;
}