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  • light_oj 1336 约数和奇偶性

    light_oj 1336  约数和奇偶性

    D - Sigma Function
    Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

    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
    ",tag++,n-ans);
        }
        return 0;
    }
    View Code

     补充:

    偶然看到的,这思路真是厉害

    http://www.hardbird.net/?p=449

    没有AC不了的题,只有不努力的ACMER!
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  • 原文地址:https://www.cnblogs.com/--560/p/4550128.html
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