LightOJ 1336 Sigma Function(唯一分解定理)

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题目:

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

$n=p_{1}^{e_{1}}\times p_{2}^{e_{2}}\times ...\times p_{k}^{e_{k}}$

Then we can write,

$\sigma(n)=\frac{p_{1}^{e_{1}+1}-1}{p_{1}-1}\times \frac{p_{2}^{e_{2}+1}-1}{p_{2}-1}\times ...\times \frac{p_{k}^{e_{k}+1}-1}{p_{k}-1}$

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 $n (1 ≤ n ≤ 10^{12})$.

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

题目链接

$\because \sigma(n)=\frac{p_{1}^{e_{1}+1}-1}{p_{1}-1}\times \frac{p_{2}^{e_{2}+1}-1}{p_{2}-1}\times ...\times \frac{p_{k}^{e_{k}+1}-1}{p_{k}-1}$，奇数$\times$奇数$=$奇数，偶数$\times$偶数$=$偶数，奇数$\times$偶数$=$偶数

$\therefore$若$\sigma$中任意一项$\frac{p^{e+1}-1}{p-1}$为偶数，则$\sigma$为偶数，只有$\sigma$中所有$\frac{p^{e+1}-1}{p-1}$都是奇数$\sigma$才是奇数

• 当$p=2$时，$\frac{p^{e+1}-1}{p-1}$一定为奇数
• 当$p\ne2$时($p$一定为奇数，因为素数中只有$2$是偶数)，当且仅当$e$为偶数时$\frac{p^{e+1}-1}{p-1}$为奇数

设$k=e+1$则$\frac{p^{e+1}-1}{p-1}=\frac{p^{k}-1}{p-1}$，若$k$为偶数($e$为奇数)则可化简为$\frac{(p^{\frac{k}{2}}-1)\times (p^{\frac{k}{2}}+1)}{p-1}$，$\because p^{\frac{k}{2}}$为奇数，$\therefore p^{\frac{k}{2}}-1$和$p^{\frac{k}{2}}+1$都为偶数，$\therefore \frac{(p^{\frac{k}{2}}-1)\times (p^{\frac{k}{2}}+1)}{p-1}=(p^{\frac{k}{2}}-1)\times \frac{p^{\frac{k}{2}}+1}{p-1}=\frac{p^{\frac{k}{2}}-1}{p-1}\times (p^{\frac{k}{2}}+1)$，无论$\frac{p^{\frac{k}{2}}+1}{p-1}$或$\frac{p^{\frac{k}{2}}-1}{p-1}$是奇数还是偶数当它乘上一个偶数$p^{\frac{k}{2}}-1$或$p^{\frac{k}{2}}+1$之后整体一定是一个偶数，若$k$为奇数则分子不可拆分

$\therefore$当且仅当$e$为偶数时$\frac{p^{e+1}-1}{p-1}$为奇数。

即对于任意一个数$x$，$\sigma(x^{2})$和$\sigma(2^{y}\times x^{2})$一定是奇数(因为当$p=2$时，$\frac{p^{e+1}-1}{p-1}$一定为奇数)。

对于题目要求出$1\rightarrow n$中$\sigma(n)$为偶数的个数，先求出$1\rightarrow n$中$\sigma(n)$为奇数的个数，分两种情况

1. 求出所有$\sigma(x^{2})(x^{2}\in [1,n])$数量，那么对于所有$i\in[1,\lfloor\sqrt{n}\rfloor]$，都有$i^{2}\in[1,n]$，数量为$\lfloor\sqrt{n}\rfloor$
2. 求出所有$\sigma(2\times x^{2})((2\times x^{2})\in [1,n])$数量，那么对于所有$i\in[1,\lfloor\sqrt{\frac{n}{2}}\rfloor]$，都有$(2\times i^{2})\in[1,n]$，数量为$\lfloor\sqrt{\frac{n}{2}}\rfloor$

$\sigma(2^{y}\times x^{2})$中当$y>1$时$\sigma(2^{y}\times x^{2})$可以被分解为$\sigma(z^{2})(z\in[1,x])$或$\sigma(2\times z^{2})(z\in[1,n])$

例如$\sigma(2^{2}\times x^{2})=\sigma((2\times x)^{2})$，$\sigma(2^{3}\times x^{2})=\sigma(2\times 2^{2}\times x^{2})=\sigma(2\times (2\times x)^{2})$

所以无需重复计算。

最后把两种数量相加即为$1\rightarrow n$中$\sigma(n)$为奇数的个数，用$n$减去$1\rightarrow n$中$\sigma(n)$为奇数的个数即为$1\rightarrow n$中$\sigma(n)$为偶数的个数。

AC代码:

#include <bits/stdc++.h>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define pb push_back
#define mp make_pair
#define lowbit(x) (x&(-x))
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<double,double> PDD;
typedef pair<ll,ll> PLL;
const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 5;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = asin(1.0) * 2;
const double e = 2.718281828459;
template <class T>
inline bool read(T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) {
        return 0;
    }
    while (c != '-' && (c < '0' || c > '9')) {
        c = getchar();
    }
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') {
        ret = ret * 10 + (c - '0');
    }
    ret *= sgn;
    return 1;
}
template <class T>
inline void out(T x) {
    if (x > 9) {
        out(x / 10);
    }
    putchar(x % 10 + '0');
}

int main(int argc, char *argv[]) {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    ll t;
    read(t);
    for (ll Case = 1, n; Case <= t; ++Case) {
        read(n);
        printf("Case %lld: %lld\n", Case, n - ll(sqrt(n)) - ll(sqrt(n / 2)));
    }
#ifndef ONLINE_JUDGE
    fclose(stdin);
    fclose(stdout);
    system("gedit out.txt");
#endif
    return 0;
}