京东幸运数求解

暴力过了80%，知道应该优化一下。现在一想，其实就是简单的求和超过了32（或者64？）就可以直接忽略该数不用算下去。

不需要暴力求解，其实你可以把它想象成一个0 1字符串，然后就行求解即可。 public static void main(String args[]) { Scanner reader = new Scanner(System.in); while (reader.hasNextLine()) { int num = Integer.parseInt(reader.nextLine()); for (int i = 0; i < num; i++) { int count = Integer.parseInt(reader.nextLine()); if (count == 1) { System.out.println(4); continue; } if (count == 2) { System.out.println(7); continue; } int a = 1; while (Math.pow(2, a) - 2 < count) { a++; } count = (int) (count - Math.pow(2, a - 1) + 2); count--; StringBuilder sb = new StringBuilder(); for (int j = 0; j < a - 1; j++) { int index = 1 << j; if ((count & index) == 0) { sb.append("4"); } else { sb.append("7"); } } System.out.println(sb.reverse().toString()); } } }

//题目是不一样的，我是2进制和10进制 import java.util.Scanner; public class Main { public static void main(String args[]) { boolean[] B = new boolean[100000 + 1]; for (int i = 1; i < 100000 + 1; i++) { int binary =g(i); int ten = f(i); if (binary == ten) B[i] = true; } Scanner cin = new Scanner(System.in); int times = cin.nextInt(); for (int i = 0; i < times; i++) { int n = cin.nextInt(); int count = 0; for (int j = 1; j <= n; j++) {{ if (B[j]) count++; } } System.out.println(count); } } private static int f(int num){ String str = String.valueOf(num); int sum = 0; for(int i = str.length() - 1;i>=0;i--){ sum+=Integer.valueOf(str.charAt(i)+""); } return sum; } private static int g(int num){ String str = Integer.toBinaryString(num); int sum = 0; for(int i = 0;i<str.length();i++){ sum+=Integer.valueOf(str.charAt(i)+""); } return sum; } }

#include <iostream> #include <string> #include <cmath> #include <vector> using namespace std; string getKthNum(long long k) {     if (k == 1) {         return "4";     } else if (k == 2) {         return "7";     }     int numPos = ceil(log(k + 2.0) / log(2.0) - 1);     long long prevNum = pow(2.0, numPos) - 2;     long long nowNum = k - prevNum;     long long halfNum = pow(2.0, numPos - 1);     string postStr = getKthNum((nowNum > halfNum ? nowNum % halfNum : nowNum) + halfNum - 2);     string prevStr = (nowNum > halfNum) ? "7" : "4";     return prevStr + postStr; } int main() {     int n;     cin >> n;     vector<string> retVec;     while (n--) {         long long k;         cin >> k;                  string ret = getKthNum(k);         retVec.push_back(ret);     }     for (int i = 0; i < retVec.size(); i++) {         cout << retVec[i] << endl;     } return 0; } 从第k个数倒推有多少位，在用递归解

我用java超时了，最后那个第1000000000，我程序用了4s跑出来