CO-PRIME
时间限制：1000 ms | 内存限制：65535 KB
难度：3
- 描写叙述
-
This problem is so easy! Can you solve it?
You are given a sequence which contains n integers a1,a2……an, your task is to find how many pair(ai, aj)(i < j) that ai and aj is co-prime.
- 输入
- There are multiple test cases.
Each test case conatains two line,the first line contains a single integer n,the second line contains n integers.
All the integer is not greater than 10^5. - 输出
- For each test case, you should output one line that contains the answer.
- 例子输入
-
3
1 2 3 - 例子输出
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3
题意：给出n个正整数，求这n个数中有多少对互素的数。
此题中，设F(d)表示n个数中gcd为d的倍数的数有多少对，f(d)表示n个数中gcd恰好为d的数有多少对，
则F(d)=∑f(n) (n % d == 0)
f(d)=∑mu[n / d] * F(n) (n %d == 0)
上面两个式子是莫比乌斯反演中的式子。
所以要求互素的数有多少对。就是求f(1)。
而依据上面的式子能够得出f(1)=∑mu[n] * F(n)。
所以把mu[]求出来，枚举n即可了。当中mu[i]为i的莫比乌斯函数。
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; const int MAXN = 1e5 + 10;
typedef long long LL;
int cnt[MAXN], pri[MAXN], num[MAXN], pri_num, mu[MAXN], vis[MAXN], a[MAXN]; void mobius(int n) //筛法求莫比乌斯函数
{
pri_num = 0;
memset(vis, 0, sizeof(vis));
vis[1] = mu[1] = 1;
for(int i = 2; i <= n; i++) {
if(!vis[i]) {
pri[pri_num++] = i;
mu[i] = -1;
}
for(int j = 0; j < pri_num; j++) {
if(i * pri[j] > n) break;
vis[i*pri[j]] = 1;
if(i % pri[j] == 0) {
mu[i*pri[j]] = 0;
break;
}
mu[i*pri[j]] = -mu[i];
}
}
} LL get(int x)
{
return (LL)x * (x-1) / 2;
} int main()
{
mobius(100005);
int n;
while(~scanf("%d",&n)) {
int mmax = 0;
for(int i = 1; i <= n; i++) {
scanf("%d",&a[i]);
mmax = max(mmax, a[i]);
}
memset(cnt, 0, sizeof(cnt));
memset(num, 0, sizeof(num));
for(int i = 1; i <= n; i++) num[a[i]]++;
for(int i = 1; i <= mmax; i++)
for(int j = i; j <= mmax; j += i)
cnt[i] += num[j];
LL ans = 0;
for(int i = 1; i <= mmax; i++)
ans += get(cnt[i]) * mu[i];
printf("%lld\n", ans);
}
return 0;
}