Acwing 1291.轻拍牛头

题意：

给定一个序列ai，问序列中其他数中有多少个数是它的约数

思路：

暴力求法会超时。O(n²)

最优解：先储存每个数的个数，遍历x，每个x的倍数加上x的个数

注：最后每个数的答案要-1(减去本身)

Code:

#pragma GCC optimize(3)
#pragma GCC optimize(2)
#include <map>
#include <set>
#include <array>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <sstream>
#include <iostream>
#include <stdlib.h>
#include <algorithm>
#include <unordered_map>

using namespace std;

typedef long long ll;
typedef pair<int, int> PII;

#define Time (double)clock() / CLOCKS_PER_SEC

#define sd(a) scanf("%d", &a)
#define sdd(a, b) scanf("%d%d", &a, &b)
#define slld(a) scanf("%lld", &a)
#define slldd(a, b) scanf("%lld%lld", &a, &b)

const int N = 1e5 + 20;
const int M = 1e6 + 20;
const int mod = 1e9 + 7;

int n;
int a[N];
int cnt[M], ans[M];

void solve()
{
    sd(n);
    int num;
    for(int i = 0; i < n; i ++){
        sd(a[i]);
        cnt[a[i]] ++;
        num = max(num, a[i]);
    }
    for(int i = 1; i <= num; i ++){
        for(int j = i; j <= num; j += i){
            ans[j] += cnt[i];
        }
    }

    for(int i = 0; i < n; i ++){
        printf("%d
", ans[a[i]] - 1);
    }
}

int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("/home/jungu/code/in.txt", "r", stdin);
    // freopen("/home/jungu/code/out.txt", "w", stdout);
    // freopen("/home/jungu/code/out.txt","w",stdout);
#endif
    // ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    // sd(T);
    while(T --){
        solve();
    }

    return 0;
}