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Description
Sigma function is an interesting function in Number Theory. It is denoted by the Greek letter Sigma (σ). This function actually denotes the sum of all divisors of a number. For example σ(24) = 1+2+3+4+6+8+12+24=60. Sigma of small numbers is easy to find but for large numbers it is very difficult to find in a straight forward way. But mathematicians have discovered a formula to find sigma. If the prime power decomposition of an integer is
Then we can write,
For some n the value of σ(n) is odd and for others it is even. Given a value n, you will have to find how many integers from 1 to n have even value of σ.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 1012).
Output
For each case, print the case number and the result.
Sample Input
4
3
10
100
1000
Sample Output
Case 1: 1
Case 2: 5
Case 3: 83
Case 4: 947
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70017#problem/D
题意：求[1,n]中约数和为偶数的数的个数
思路：根据算术基本定理的约数和公式f(n)=(1+p1+p1^2+...+p1^k1)(1+p2+p2^2+...+p2^k2)...(1+pn+pn^3+...+pn^kn);
其中n=(p1^k1)*(p2^k2)*...*(pn^kn);(分解素因数)
由于奇数*奇数还是奇数，奇数*偶数或者偶数相乘是偶数。而素数除了2都是奇数。
f(n)的奇偶性取决于每个因子的奇偶性，只要出现一个因子是偶数时f(n)为偶数。
对每个因子，1+p+p^2+...+p^k的奇偶性，
p为素数，当p是2时，式子为偶数；
当p不是2时，p是奇数数，p^i为奇数，式子总共k+1项，即k+1个奇数的和，k为偶数时式子为奇数。
以此条件反推dfs出所有的f(n)为奇数的n，方法是每次乘以奇素数素数的两倍或者2，得到的n就是f(n)为奇数的n了。
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<vector> #include<stack> #include<queue> #include<set> #include<map> #include<string> #include<math.h> #include<cctype> using namespace std; typedef long long ll; const int maxn=1000100; const int INF=(1<<29); const double EPS=0.0000000001; const double Pi=acos(-1.0); ll n; vector<ll> odd; bool isprime[maxn]; vector<int> prime; const ll M=1000000000010; void play_prime() { memset(isprime,1,sizeof(isprime)); isprime[1]=0; for(int i=2;i<maxn;i++){ if(!isprime[i]) continue; for(int j=i*2;j<maxn;j+=i){ isprime[j]=0; } } for(int i=1;i<maxn;i++){ if(isprime[i]) prime.push_back(i); } } void dfs(int dep,ll cur) { odd.push_back(cur); if(dep>=(int)prime.size()) return; for(int i=dep;i<prime.size();i++){ ll t=prime[i]; if(t==2){ if(cur<=M/2) dfs(i,cur*2); else return; } else{ if(cur<=M/(t*t)) dfs(i,cur*t*t); else return; } } } int main() { int T;cin>>T; play_prime(); dfs(0,1); sort(odd.begin(),odd.end()); int tag=1; while(T--){ cin>>n; ll ans=upper_bound(odd.begin(),odd.end(),n)-odd.begin(); printf("Case %d: %lld\n",tag++,n-ans); } return 0; }
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原文地址：http://www.cnblogs.com/--560/p/4550128.html
