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Every positive integer can be expressed by multiplication of prime integers. Duoxida says an integer is a yada number if the total amount of 2,3,5,7,11,13 in its prime factors is even.
For instance, 18=2 * 3 * 3 is not a yada number since the sum of amount of 2, 3 is 3, an odd number; while 170 = 2 * 5 * 17 is a yada number since the sum of amount of 2, 5 is 2, a even number that satifies the definition of yada number.
Now, Duoxida wonders how many yada number are among all integers in [1,n].
The first line contains a integer T(no more than 50) which indicating the number of test cases. In the following T lines containing a integer n. ()
For each case, output the answer in one single line.
2
18
21
9
11
??????????
题意：
给你一个n，问你1到n里面有多少个数满足 因子中是2,3,5,7,11,13的个数为偶数个
题解：
预处理出所有的x，满足x只含有2,3,5,7,11,3这几个质因子，且数目为偶数。x的数目略大于10000
注意加入0个的情况，即1.
对于一个数n，枚举所有的x，对于一个x，f(n/x)即求出[1,n/x]中不含有2,3,5，7,11,13作为因子的数有多少个，这个是经典的容斥问题。
最后对所有的f(n/x)求和即可
#include<bits/stdc++.h> using namespace std; const int N = 3e6+20, M = 1e6+10, mod = 1e9+7,inf = 1e9; typedef long long ll; const ll maxn = 1e9; int cnt = 0, ans,n; ll b[N]; int a[] = {2,3,5,7,11,13}; ll gcd(ll a,ll b) {return b==0?a:gcd(b,a%b);} void dfs(ll x,int f,int num) { if(num==6) { if(!f) b[cnt++] = x; return ; } while(x<=maxn) { dfs(x,f,num+1); x*=a[num]; f^=1; } } void init() { dfs(1,0,0); sort(b,b+cnt); } void inclu(int i,int num,ll tmp) { if(tmp>n) return ; if(i>=6) { if(num==0) ans = 0; else { if(num&1) ans = ans+n/tmp; else ans = ans-n/tmp; } return ; } inclu(i+1,num,tmp); inclu(i+1,num+1,tmp*a[i]/gcd(tmp,a[i])); } void solve() { int Ans = 0; scanf("%d",&n); int tm = n; for(int i=0;i<cnt&&b[i]<=tm;i++) { n = tm/b[i]; ans = 0; inclu(0,0,1); Ans+=(n - ans); } printf("%d\n",Ans); } int main() { int T; cnt = 0; init(); scanf("%d",&T); while(T--) { solve(); } return 0; }
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原文地址：http://www.cnblogs.com/zxhl/p/5475495.html
