86 lines
2.0 KiB
C++
86 lines
2.0 KiB
C++
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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//约数和定理
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//https://baike.baidu.com/item/%E7%BA%A6%E6%95%B0%E5%92%8C%E5%AE%9A%E7%90%86/3808428
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/**
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例题：正整数360的所有正约数的和是多少？
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解：将360分解质因数可得
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360=2^3*3^2*5^1
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由约数和定理可知，360所有正约数的和为
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(2^0+2^1+2^2+2^3)×(3^0+3^1+3^2)×(5^0+5^1)=(1+2+4+8)(1+3+9)(1+5)=15×13×6=1170
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可知360的约数有1、2、3、4、5、6、8、9、10、12、15、18、20、24、30、36、40、45、60、72、90、120、180、360；
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则它们的和为1+2+3+4+5+6+8+9+10+12+15+18+20+24+30+36+40+45+60+72+90+120+180+360=1170
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*/
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const int N = 1010;
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int a[N];
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int cnt;
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//下面是求所有约数的方法
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void ben(int n) {
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for (int i = 1; i <= sqrt(n); i++) {
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if (n % i == 0) {
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a[++cnt] = i;
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if (n != i * i)
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a[++cnt] = n / i;
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}
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}
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}
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/**
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* 功能：快速幂
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* @param m
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* @param k
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* @return
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*/
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int qmi(int m, int k) {
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int res = 1, t = m;
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while (k) {
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if (k & 1) res = res * t;
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t = t * t;
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k >>= 1;
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}
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return res;
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}
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/**
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* C++版本的约数和定理
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* https://blog.csdn.net/qq_40924940/article/details/86525912
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* @param n
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* @return
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*/
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LL getSum(LL n) {
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LL res = 1;
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for (LL i = 2; i * i <= n; i++) {
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LL k = 0;
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while (n % i == 0) {
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n = n / i;
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k++;
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}
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res = res * ((1 - qmi(i, k + 1)) / (1 - i));
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}
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if (n != 1) res *= (1 + n);
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return res;
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}
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int main() {
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//1、算出所有约数再求和
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ben(360);
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//排序一下，否则顺序有点看不懂
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sort(a + 1, a + 1 + cnt);
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for (int i = 1; i <= cnt; i++)cout << a[i] << " ";
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cout << endl;
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//计算一下总和
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long long sum = 0;
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for (int i = 1; i <= cnt; i++)sum += a[i];
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cout << sum << endl;
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//2、根据公式算一下约数和
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cout << getSum(360) << endl;
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return 0;
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}
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