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package class_2022_05_1_week;
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// 来自学员问题，蓝桥杯练习题
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// f(i) : i的所有因子，每个因子都平方之后，累加起来
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// 比如f(10) = 1平方 + 2平方 + 5平方 + 10平方 = 1 + 4 + 25 + 100 = 130
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// 给定一个数n，求f(1) + f(2) + .. + f(n)
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// n <= 10的9次方
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// O(n)的方法都会超时！低于它的！
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// O(根号N)的方法，就过了，一个思路
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// O(log N)的方法，
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public class Code04_SumOfQuadraticSum {
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// 暴力方法
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public static long sum1(long n) {
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int[] cnt = new int[(int) n + 1];
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for (int num = 1; num <= n; num++) {
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for (int j = 1; j <= num; j++) {
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if (num % j == 0) {
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cnt[j]++;
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}
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}
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}
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long ans = 0;
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for (long i = 1; i <= n; i++) {
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ans += i * i * (long) (cnt[(int) i]);
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}
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return ans;
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}
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// 正式方法
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// 时间复杂度O(开平方根N + 开平方根N * logN)
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public static long sum2(long n) {
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// 100 -> 10
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// 200 -> 14
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long sqrt = (long) Math.pow((double) n, 0.5);
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long ans = 0;
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for (long i = 1; i <= sqrt; i++) {
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ans += i * i * (n / i);
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}
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// 后半段
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// 给你一个个数，二分出几个因子，处在这个个数上！
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// 由最大个数(根号N), 开始二分
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for (long k = n / (sqrt + 1); k >= 1; k--) {
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ans += sumOfLimitNumber(n, k);
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}
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return ans;
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}
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// 平方和公式n(n+1)(2n+1)/6
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public static long sumOfLimitNumber(long v, long n) {
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long r = cover(v, n);
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long l = cover(v, n + 1);
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return ((r * (r + 1) * ((r << 1) + 1)
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- l * (l + 1) * ((l << 1) + 1)) * n)
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/ 6;
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}
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public static long cover(long v, long n) {
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long l = 1;
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long r = v;
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long m = 0;
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long ans = 0;
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while (l <= r) {
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m = (l + r) / 2;
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if (m * n <= v) {
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ans = m;
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l = m + 1;
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} else {
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r = m - 1;
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}
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}
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return ans;
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}
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// 实验
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// 解法来自观察
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// 打表（暴力）
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// f(1) + ... + f(n)
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public static void test(int n) {
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int[] cnt = new int[n + 1];
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for (int num = 1; num <= n; num++) {
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for (int j = 1; j <= num; j++) {
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if (num % j == 0) {
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cnt[j]++;
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}
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}
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}
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for (int i = 1; i <= n; i++) {
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System.out.println("因子 : " + i + ", 个数 : " + cnt[i]);
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}
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}
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public static void main(String[] args) {
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// test(100);
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System.out.println("测试开始");
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for (long i = 1; i < 1000; i++) {
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if (sum1(i) != sum2(i)) {
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System.out.println("出错了!");
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}
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}
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System.out.println("测试结束");
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long n = 50000000000L; // 5 * 10的10次方
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long start = System.currentTimeMillis();
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sum2(n);
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long end = System.currentTimeMillis();
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System.out.println("大样本测试，n = " + n);
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System.out.println("运行时间 : " + (end - start) + " ms");
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}
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}
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