[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。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<cctype>
 8 #include<vector>
 9 #include<stack>
10 #include<queue>
11 using namespace std;
12 #define enter printf("\n")
13 #define space printf(" ")
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const int eps = 1e-8;
19 const int maxn = 1e5 + 5;
20 const int max_size = 1e6 + 5;
21 inline ll read()
22 {
23     ll ans = 0;
24     char ch = getchar(), last = ' ';
25     while(!isdigit(ch)) {last = ch; ch = getchar();}
26     while(isdigit(ch))
27     {
28         ans = ans * 10 + ch - '0'; ch = getchar();
29     }
30     if(last == '-') ans = -ans;
31     return ans;
32 }
33 inline void write(ll x)
34 {
35     if(x < 0) x = -x, putchar('-');
36     if(x >= 10) write(x / 10);
37     putchar(x % 10 + '0');
38 }
39 
40 int n, a[maxn];
41 int num[max_size], Max = 0, ans[max_size];
42 
43 int main()
44 {
45     n = read();
46     for(int i = 1; i <= n; ++i) 
47     {
48         a[i] = read();
49         Max = max(Max, a[i]);
50         num[a[i]]++;        //开一个桶 
51     }
52     for(int i = 1; i <= Max; ++i) if(num[i])
53         for(int j = i; j <= Max; j += i) ans[j] += num[i];
54     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 
55     return 0;
56 }

View Code