CF GYM 100548 Last Defence (2014ACM Xi ' an live game problem K)

Problemk. Last defence

Description

Given Integersa and B. Sequence S is defined as follow:

? S0 = A

? S1 = B

? Si = | Si?1? si?2| For i≥2

Count the number ofdistinct numbers in S.

Input

The first line ofthe input gives the number of test cases, T. t test cases follow. Tis about 100000.

Each test caseconsists of one line-two space-separated integers A, B. (0≤a, b≤1018).

Output

For each test case,output-one line containing ' case #x: Y ', where x is the ' Test Casenumber (starting from 1) and Y-is the Number of distinct numbers inS.

Samples

Sample Input	Sample Output
2 7 4 3 5	case #1:6 Case #2:5

Knowledge Points:

Ad-hoc, the method of dividing.

Main topic:

Given series The first two items of S, requirements after the various meet si= | Si?1? si?2| (The absolute value of the first two differences). Ask the number of different numbers in the whole series S.

Problem Solving Ideas:

first, it is easy to see that when I was big enough, I would eventually seethe repetition of "xx0xx0 ...". So the number of different numbers must be limited.

the reason, for the numberYand theX,Ymust be able to writeKx+bIn the process of generating a sequence of numbers, it appearsKx+b,X,(k-1) X+b,(k-2) X+b,X,...,2x+b,X,X+b,B,X, the number of different numbers that appear is(kx+b)/x, then the problem becomes a numberXand theBproblem, it can be found that this is the process of dividing the method. Every time I make a divideGCD (x, y), the number of different numbers is moreX/Y. At the end, add a0.

There are some special cases to consider, such as the number 0.

Reference code:

#include <iostream>using namespace Std;long long A, B, Ans;int ncase, Ccase;long long calc (Long long A, long long B) {    long long ret = 0;    while (b) {        long long t = B;        RET + = A/b;        b = a% B;        A = t;    }    return ret + 1;} int main () {    ios::sync_with_stdio (false);    Cin >> Ncase;    while (ncase--) {        cin >> a >> b;        if (a = = 0 && b = = 0) {            ans = 1;        } else if (a = = 0 | | b = = 0) {            ans = 2;        } else {            ans = c ALC (A, b);        cout << "Case #" << ++ccase << ":" << ans << endl;    }    return 0;}

CF GYM 100548 Last Defence (2014ACM Xi ' an live game problem K)