Sigma Function
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).
OutputFor each case, print the case number and the result.
Sample Input4
3
10
100
1000
Sample OutputCase 1: 1
Case 2: 5
Case 3: 83
Case 4: 947
题意：函数σ(n)等于n的所有因数之和，问满足σ(k)为偶数的k(1<=k<=n)有多少个。
分析：题目给出求σ(n)的公式σ(n)=((p1^(e1+1)-1)/(p2-1))*((p2^(e2+1)-1)/(p2-1))*......*((pk^(ek+1)-1)/(pk-1)),
又根据等比数列公式(p^(e+1)-1)/(p-1)=1+p^1+p^2+......+p^e，
我们可以先求σ(k)为奇数的个数ans再用n减去ans便是σ(k)为偶数的个数：
当σ(n)为奇数时，(p^(e+1)-1)/(p-1)=1+p^1+p^2+......+p^e 必为奇数，
(1) 当p为偶数的时候，因为既是偶数又是素数的只有2，故p=2，
此时(p^(e+1)-1)/(p-1)=2^(e+1)-1必为奇数；
(2)当p为奇数时，p^k必为奇数，要使(p^(e+1)-1)/(p-1)=1+p^1+p^2+......+p^e为奇数，
e+1必为奇数，则e必为偶数；
因此，只有p1,p2,......pk满足(1)或(2)的时候σ(k)为奇数。
由算术基本定理知，n=p1^e1 * p2^e2 * ...... * pk^ek ，
1.当n满足条件(2)时，因为ei为偶数，所以n必为完全平方数，而且可以证明完全平方数n的σ(n)必为奇数；
2.当n满足条件(1)时，n=2^t * p1^e1 * p2^e2 * ...... * pk^ek ，若t为偶数则它n就是完全平方数，即上面的1，
若t为奇数则n=2*X (X为完全平方数)。
因此，只要排除掉n以内的所有完全平方数X和2*X，就可以得到σ(k)为偶数的个数。
#include<cstdio> #include<cmath> int main() { int T,cas=0; long long N,ans; scanf("%d",&T); while(T--) { scanf("%lld",&N); ans=(long long)sqrt(N)+(long long)sqrt(N/2.0); printf("Case %d: %lld\n",++cas,N-ans); } return 0; }