洛谷[USACO08DEC] Patting Heads S
题目传送门
题目难度：普及/提高一
题面翻译
今天是贝茜的生日，为了庆祝自己的生日，贝茜邀你来玩一个游戏。
贝茜让 N N N ( 1 ≤ N ≤ 1 0 5 1\leq N\leq 10^5 1≤N≤105) 头奶牛坐成一个圈。除了 1 1 1 号与 N N N 号奶牛外， i i i 号奶牛与 i − 1 i-1 i−1 号和 i + 1 i+1 i+1 号奶牛相邻。 N N N 号奶牛与 1 1 1 号奶牛相邻。农夫约翰用很多纸条装满了一个桶，每一张包含了一个不一定是独一无二的 1 1 1 到 1 0 6 10^6 106 的数字。
接着每一头奶牛 i i i 从桶中取出一张纸条 A i A_i Ai。每头奶牛轮流走上一圈，同时拍打所有手上数字能整除在自己纸条上的数字的牛的头，然后坐回到原来的位置。牛们希望你帮助他们确定，每一头奶牛需要拍打的牛的数量。
题目描述
It’s Bessie’s birthday and time for party games! Bessie has instructed the N (1 <= N <= 100,000) cows conveniently numbered 1…N to sit in a circle (so that cow i [except at the ends] sits next to cows i-1 and i+1; cow N sits next to cow 1). Meanwhile, Farmer John fills a barrel with one billion slips of paper, each containing some integer in the range 1…1,000,000.
Each cow i then draws a number A_i (1 <= A_i <= 1,000,000) (which is not necessarily unique, of course) from the giant barrel. Taking turns, each cow i then takes a walk around the circle and pats the heads of all other cows j such that her number A_i is exactly
divisible by cow j’s number A_j; she then sits again back in her original position.
The cows would like you to help them determine, for each cow, the number of other cows she should pat.
输入格式
* Line 1: A single integer: N
* Lines 2…N+1: Line i+1 contains a single integer: A_i
输出格式
* Lines 1…N: On line i, print a single integer that is the number of other cows patted by cow i.
样例 #1
样例输入 #1
5
2
1
2
3
4
样例输出 #1
2
0
2
1
3
提示
The 5 cows are given the numbers 2, 1, 2, 3, and 4, respectively.
The first cow pats the second and third cows; the second cows pats no cows; etc.
题目分析：由于本题纯暴力枚举会炸的，优化解法：统计每个数字 Ai 出现的次数。对于每头奶牛 i枚举 Ai 的所有因数 并统计出现的次数。时间复杂度：O(NM)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e6 + 10;int ans[N],cnt[N],a[N];
int n;ll read()
{ll s=0,f=1;char ch=getchar();while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}while (ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}return s*f;
}int main() {n = read();for(int i = 1; i <= n; i++) {a[i] = read();cnt[a[i]]++;}int t = 0;for(int i = 1; i <= n; i++){t = 0;for(int j = 1; j <= a[i] / j; j++){if(a[i] % j == 0) {t += cnt[j];if(j != a[i] / j) t += cnt[a[i] / j];}} ans[i] = t - 1;}for(int i = 1; i <= n; i++) cout<<ans[i]<<endl;return 0;
}