CF-60D - Savior(并查集+数论)
Description
Misha decided to help Pasha and Akim be friends again. He had a cunning plan — to destroy all the laughy mushrooms. He knows that the laughy mushrooms can easily burst when they laugh. Mushrooms grow on the lawns. There are a[t] mushrooms on the t-th lawn.
Misha knows that the lawns where the mushrooms grow have a unique ability. A lawn (say, i) can transfer laugh to other lawn (say, j) if there exists an integer (say, b) such, that some permutation of numbers a[i], a[j] and b is a beautiful triple (i ≠ j). A beautiful triple is such three pairwise coprime numbers x, y, z, which satisfy the following condition: x2 + y2 = z2.
Misha wants to know on which minimal number of lawns he should laugh for all the laughy mushrooms to burst.
Input
The first line contains one integer n (1 ≤ n ≤ 106) which is the number of lawns. The next line contains n integers ai which are the number of mushrooms on the i-lawn (1 ≤ ai ≤ 107). All the numbers are different.
Output
Print a single number — the minimal number of lawns on which Misha should laugh for all the mushrooms to burst.
Sample Input
1 2
1
2 1 2
2
2 3 5
1
思路:所有勾股数都可以表示为x^2-y^2,,2*x*y,,,x^2+y^2 枚举所有勾股数,然后把能合并的合并,最后查看有几个不同的集合就是答案。
#include<iostream> #include<cstring> #include<cstdio> using namespace std; const int mm=1e7+9; int root[mm],f[mm],n,num; int look(int x) { if(x^root[x]) root[x]=look(root[x]); return root[x]; } void uni(int a,int b) { a=look(a);b=look(b); if(root[a]!=0&&root[b]!=0&&a!=b) { root[a]=b;++num; } } int gcd(int a,int b) { int z; while(a) { z=a;a=b%a;b=z; } return b; } int main() { long long x,y,z; while(scanf("%d",&n)!=EOF) { memset(root,0,sizeof(root)); for(int i=0;i<n;++i) {scanf("%d",&f[i]);root[f[i]]=f[i];} num=0; for(long long i=1;i<mm;i++) { for(long long j=i+1;2*i*j<mm;j+=2) { x=j*j-i*i;y=2*i*j;z=i*i+j*j; if(x>mm||y>mm)break; if(gcd(i,j)!=1)continue; uni(x,y); if(z<mm) uni(x,z); if(z<mm) uni(y,z); } } printf("%d\n",n-num); } }