最大公因数的一个小性质
最大公因数存在以下的性质
设s = gcd(a,b)，则s = gcd(xa+yb,ma+nb);
其中，x,y,m,n为正整数
这里我们通过一个程序来进行说明
#include<bits/stdc++.h>
#define ll long long
using namespace std;
ll aim[100];
ll gcd(ll a,ll b)
{
if(a<b)return gcd(b,a);
return (b==0)? a:gcd(b,a%b);
}
int main()
{
aim[0]=28;
aim[1]=56;
for(int i=2;i<100;i++)
{
aim[i]=aim[i-1]+aim[i-2];
}
cout<<"We get the biggest common num is "<<gcd(aim[0],aim[1])<<endl;
int a,b;
cout<<"===============OK==============="<<endl;
while(cin>>a>>b)
{
cout<<"You slect the num is "<<aim[a]<<" and "<<aim[b]<<endl;
cout<<"So the biggest common num is "<<gcd(aim[a],aim[b])<<endl;
cout<<"---------------"<<endl;
}
return 0;
}
上述程序其实是对已知的28,56按照斐波那契数列的原理进行计算
之后我们随意输入a,b来检验数组中a,b位置的两个数的最大公因数
输入几组后，我们发现其得到的结果均为gcd(a,b)
比如，输入数据
2 3
23 64
53 32
26 36
得到的结果如下
We get the biggest common num is 28
===============OK===============
2 3
You slect the num is 84 and 140
So the biggest common num is 28
---------------
23 64
You slect the num is 2100700 and 777780920988064
So the biggest common num is 28
---------------
53 32
You slect the num is 3908348148460 and 159680836
So the biggest common num is 28
---------------
26 36
You slect the num is 8898708 and 1094468732
So the biggest common num is 28
---------------
^Z
--------------------------------
Process exited after 8.104 seconds with return value 0
下面我我给出一个例题
官方给出的解题报告如下
Problem A
观察题目可知两个容器按照如下状态转移
(F1,F2)->(F2,F3)...(FN+1,FN+1+Y)
之后继续按FN=FN-1+FN-2递推，易知这个序列分成了两部分，1~FN和FN后面的
将后面的那个序列看成新的序列{A} 若求GCD（FM+1,FM+2）
则是在求GCD（AM-N+1,AM-N+2）
由更相减损术知：GCD(AM-N+2,AM-N+1)=GCD(AM-N+1,AM-N+2-AM-N+1)=GCD(AM-N+1,AM-N)...=GCD(A1,A2)
A1=FN+1
A2=FN+2+Y
M<70可以直接递推求解。
Std:
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {
if (b == 0)return a;
else return gcd(b, a % b);
}
ll F[123];
ll b1,b2;
int main() {
int t;
int testCase = 0;
cin >> t;
while (t--) {
ll x, n, y, m;
cin >> x >> n >> y >> m;
a[1] = 0, a[2] = x;
for (int i = 3; i <= 72; i++) {
F[i] = F[i - 1] + F[i - 2];
}
b1 = a[n + 1];
b2 = a[n + 2]+y;
printf("Case %d: %lld\n", ++testCase, gcd(b1, b2));
}
}
OK
