• LightOJ 13361336


    http://lightoj.com/volume_showproblem.php?problem=1336

    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

     题目大意:给你一个数n,让你求从1到n中因子和为偶数的数共有多少个,可以用唯一分定理的公式

    来找(我这样写过,不过超时)

    那么我们可以通过超时的代码将100中不满足的数打出来,如下:

    2 3

    1 1
    4 7
    8 15
    9 13
    16 31
    18 39
    25 31
    32 63
    36 91
    49 57
    50 93
    64 127
    72 195
    81 121
    98 171
    100 217

    发现100中,因子和为奇数的有:

    1 2 4 6 9 16 18 25 32 36 49 50 64 72 81 98 100

    有没有发现,这些数字有一个规律,他们是 x^2, 2*x, 2*x^2, 只要100中的数满足这三个中的一个,那么,这个数就是不满足的,

    总数-不满足的个数 = 满足的个数

    我们还可以发现:当x为偶数时2*x和x^2会有重复的部分

                            当x为奇数时2*x和2*x^2会有重复的部分  

    那么我们可以将2*x省去,我们求求出x^2的个数和2*x^2的个数,然后用总数减去它们的个数即可

    AC代码:

    #include<stdio.h>
    #include<math.h>
    #include<string.h>
    #include<stdlib.h>
    #include<algorithm>
    
    using namespace std;
    typedef long long ll;
    const int N = 1e6 + 10;
    
    int main()
    {
        int t, p = 0;
        ll n;
        scanf("%d", &t);
        while(t--)
        {
            p++;
            ll x, y;
            scanf("%lld", &n);
            x = (ll)sqrt(n);//计算n中x^2的个数
            y = (ll)sqrt(1.0 * n / 2);//计算n中2*x^2的个数
            printf("Case %d: %lld
    ", p, n - x - y);
        }
        return 0;
    }

    TLE代码(可以用来打表找规律):

    #include<stdio.h>
    #include<math.h>
    #include<string.h>
    #include<stdlib.h>
    #include<algorithm>
    
    using namespace std;
    typedef long long ll;
    const int N = 1e6 + 10;
    
    int prime[N], k;
    bool Isprime[N];
    
    void Prime()
    {
        k = 0;
        memset(Isprime, true, sizeof(Isprime));
        Isprime[1] = false;
        for(int i = 2 ; i < N ; i++)
        {
            if(Isprime[i])
            {
                prime[k++] = i;
                for(int j = 2 ; j * i < N ; j++)
                    Isprime[j * i] = false;
            }
        }
    }
    
    ll solve(ll n)
    {
        ll x, y, a;
        ll num = 1;
        for(ll i = 0 ; i < k && prime[i] * prime[i] <= n ; i++)
        {
            x = 0;
            if(n % prime[i] == 0)
              {
                  while(n % prime[i] == 0)
                  {
                      x++;
                      n /= prime[i];
                  }
                  x += 1;
                  a = 1;
                  for(ll j = 1 ; j <= x ; j++)
                  a *= prime[i];
                  y = a - 1;
                   num *= y /(prime[i] - 1);
              }
        }
        if(n > 1)
            num *= ((n * n - 1) / (n - 1));
        return num;
    }
    
    int main()
    {
        Prime();
        int t, x = 0;
        ll n;
        scanf("%d", &t);
        while(t--)
        {
            x++;
            scanf("%lld", &n);
            ll num = 0;
            for(ll i = 2 ; i <= n ; i++)
            {
                ll m = solve(i);
                if(m % 2 != 0)
                   {
                       printf("%lld %lld
    ", i, m);
                       num++;
                   }
            }
            printf("Case %d: %lld
    ", x, n - 1 - num);
        }
        return 0;
    }
     
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  • 原文地址:https://www.cnblogs.com/qq2424260747/p/4930861.html
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