The 2014 ACM-ICPC Asia Regional Contest Xi'an Site K ast Defence

ProblemK. Last Defence

Description

Given two 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

题目大意：给定两个数，每次由两个数的差得到一个新的数，求一共能得到几个不同的数。

解题思路：因为最终会得到X X 0这样循环出现的数，所以不同数字个数是有限个的。因为数字太大，直接求会T，正解是辗转相除法。我的方法是观察找规律，根据两个数的差值和两个数之间的关系，可以得到后面数字的个数。

代码如下：

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#define ll long long#define mod=1e9+7; using namespace std;const int maxn=10000005;ll ans;void getans(ll a,ll b){    if(a==b)        return ;    ll c=a-b;   //根据a,c,b之间的关系可以得到一段长度为cur的等差数列，长度即数字个数    ll cur=a/c;   //得到多段等差数列    a=a-(cur-1)*c;    ans+=cur-1;    getans(a,c);}int main(void){    int t;    ll x,y,a,b,c;    scanf("%d",&t);    for(int ca=1;ca<=t;ca++)    {        ans=0;        scanf("%I64d%I64d",&x,&y);        if(x==y)   //x,y相同时特判一下        {            if(x==0)                printf("Case #%d: 1\n",ca);            else                printf("Case #%d: 2\n",ca);            continue;        }        if(x>y)   //保证得到的a,b值是a>b        {            ll c=x-y;            a=x;            if(c>y)                b=c;            else                b=y;            }        else        {            ll c=y-x;            a=y;            if(c>x)                b=c;            else                b=x;        }        getans(a,b);         printf("Case #%d: %I64d\n",ca,ans+2);    }   }

正解辗转相除法：

#include <cstdio>#include <cstring>#define ll long long using namespace std;ll ans;void get_ans(ll a,ll b){    if(b==0)        return;    ans+=a/b;    get_ans(b,a%b);}int main(){    int t;    ll a,b;    scanf("%d",&t);    for(int ca=1;ca<=t;ca++)    {        ans=0;        scanf("%I64d%I64d",&a,&b);        if(a==0||b==0)        {                 if(a==0&&b==0)                       printf("Case #%d: 1\n",ca);            else                printf("Case #%d: 2\n",ca);            continue;        }        if(a>b)            get_ans(a,b);        else            get_ans(b,a);        printf("Case #%d: %I64d\n",ca,ans+1);    }    return 0;}