hdu5072 2014 Asia AnShan Regional Contest C Coprime

最后一次参加亚洲区……

题意：给出n(3 ≤ n ≤ 105)个数字，每个数ai满足1 ≤ ai ≤ 105，求有多少对(a,b,c)满足[(a, b) = (b, c) = (a, c) = 1] or [(a, b) ≠ 1 and (a, c) ≠ 1 and (b, c) ≠ 1]，都互素或都不互素。

思路：如果是两个数，互素比较好求，边遍历边分解质因子，利用容斥原理即可知道前面与自己互素的有多少。left_prime[i]表示左边与自己互素的，left_no_prime[i]表示左边与自己不互素的数量，同理right表示右边的情况。

总用情况减去不合法情况即可，总共情况C(n,3)，非法情况有以下几种：

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;

const int MAXN = 100011;
long long prime[MAXN+10];
int getPrime(){
memset(prime,0,sizeof(prime));
for(int i=2;i<=MAXN;i++){
if(!prime[i]) prime[++prime[0]]=i;
for(int j=1;j<=prime[0]&&prime[j]<=MAXN/i;j++){
prime[prime[j]*i]=1;
if(i%prime[j]==0) break;
}
}
return prime[0];
}

int Stack[MAXN][10], s_top[MAXN];
int arr[MAXN],num[MAXN];
int left_prime[MAXN],left_no_prime[MAXN];
int right_prime[MAXN],right_no_prime[MAXN];

void factor_full_Stack(){
for(int i = 1;i <= 100000;i ++){
s_top[i] = 0;
int x = i;
for(int j = 1;x != 1;j ++){
if(prime[j] * prime[j] > i){
Stack[i][s_top[i] ++] = x;
break;
}
if(x % prime[j] == 0) Stack[i][s_top[i] ++] = prime[j];
while(x % prime[j] == 0) x /= prime[j];
}
}
return ;
}

void factor(int x){
int end = (1 << s_top[x]);
for(int i = 1;i < end;i ++){
int tmp = 1;
for(int j = 0;j < s_top[x];j ++){
if(i & (1 << j)){
tmp *= Stack[x][j];
}
}
num[tmp] ++;
}
return ;
}

int cal_coprime_num(int x){
int res = 0;
int end = (1 << s_top[x]);
for(int i = 1;i < end;i ++){
int cnt = 0,tmp = 1;
for(int j = 0;j < s_top[x];j ++){
if(i & (1 << j)){
cnt ++;
tmp *= Stack[x][j];
}
}
if(cnt & 1){
res += num[tmp];
}else{
res -= num[tmp];
}
}
return res;
}

void init(){
getPrime();
factor_full_Stack();
}
int main(){
init();
int T,n;
scanf("%d",&T);
while(T --){
scanf("%d",&n);
for(int i = 0;i < n;i ++){
scanf("%d",&arr[i]);
}

memset(num, 0, sizeof(num));
for(int i = 0;i < n;i ++){
left_no_prime[i] = cal_coprime_num(arr[i]);
left_prime[i] = i  - left_no_prime[i];
factor(arr[i]);
}

memset(num, 0, sizeof(num));
for(int i = n - 1;i >= 0;i --){
right_no_prime[i] = cal_coprime_num(arr[i]);
right_prime[i] = n - i - 1 - right_no_prime[i];
factor(arr[i]);
}

long long res = (long long)n * (n-1) *(long long)(n-2) / 6;
long long cha = 0;
for(int i = 0;i < n;i ++){
cha += (long long) left_prime[i] * left_no_prime[i];
cha += (long long) left_prime[i] * right_no_prime[i];
cha += (long long) right_prime[i] * left_no_prime[i];
cha += (long long) right_prime[i] * right_no_prime[i];
}
res -= cha / 2;
printf("%I64d\n",res);
}
return 0;
}

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