[USACO08DEC]Patting Heads

嘟嘟嘟

这题还是比较水的。首先O(n²)模拟显然过不了，那就换一种思路，考虑每一个数对答案的贡献，显然一个数a[i]会对后面的a[i] * 2, a[i] * 3，a[i] * 4……都贡献1,。那么就想线性求因数个数一样，对于每一个a[i]，都计算出对能被他整出的数的贡献。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter printf("
")
#define space printf(" ")
#define Mem(a) memset(a, 0, sizeof(a))
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const int eps = 1e-8;
const int maxn = 1e5 + 5;
const int max_size = 1e6 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch))
    {
        ans = ans * 10 + ch - '0'; ch = getchar();
    }
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int n, a[maxn];
int num[max_size], Max = 0, ans[max_size];

int main()
{
    n = read();
    for(int i = 1; i <= n; ++i) 
    {
        a[i] = read();
        Max = max(Max, a[i]);
        num[a[i]]++;        //开一个桶 
    }
    for(int i = 1; i <= Max; ++i) if(num[i])
        for(int j = i; j <= Max; j += i) ans[j] += num[i];
    for(int i = 1; i <= n; ++i) write(ans[a[i]] - 1), enter;    //因为还有他自己，所以要减1 
    return 0;
}