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| #include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
const int MAXN = 1010;
int GCD(int a,int b)
{
if(b == 0)
return a;
return GCD(b,a%b);
}
int A[MAXN],B[MAXN],C[MAXN*10];
int main()
{
int N;
while(~scanf("%d",&N))
{
for(int i = 0; i < N; ++i)
scanf("%d",&A[i]);
memset(B,0,sizeof(B));
memset(C,0,sizeof(C));
B[0] = 1;
int a,b,gcd,fm = 0,k;
for(int i = 1; i < N; ++i)
{
if(A[i] != A[0])
{
b = A[i]*A[0];
a = abs(A[i]-A[0])*2;
gcd = GCD(a,b);
a /= gcd;//约分
b /= gcd;//约分
fm = GCD(a,fm);//迭代求n个分母最大公约数
for(int j = 2; b > 1; ++j)//分解质因数
{
if(b % j == 0)
{
k = 0;
while(b % j == 0){
b /= j;
k++;
}
if(k > C[j]) //C[] 数组记录素因子的幂 对应的最大值
C[j] = k;
}
}
}
}
int tmp;
for(int i = 0; i < MAXN*10; ++i)//大数乘法迭代求n个数最小公倍数
{
for(int j = 0; j < C[i]; ++j)
{
tmp = 0;
for(int k = 0; k < MAXN; ++k)
{
B[k] = B[k]*i + tmp;
tmp = B[k] / 10000;
B[k] %= 10000;
}
}
}
int i = 999;
while(i > 0 && B[i] == 0)
i--;
printf("%d-",B[i]);
for(--i; i >= 0; --i)
printf("%04d+",B[i]);
printf(" %d,\n",fm);
}
return 0;
}
|