[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("\n")
 #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-;
 const int maxn = 1e5 + ;
 const int max_size = 1e6 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch))
     {
         ans = ans *  + ch - ''; ch = getchar();
     }
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

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

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