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

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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;

}

View Code

 补充：

偶然看到的，这思路真是厉害

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