LightOJ 13361336

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

Sigma Function

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

Submit Status Practice LightOJ 1336

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;
}