(Problem 92)Square digit chains
A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
For example,
44 32 13 10 1 1 85 89 145 42 20 4 16 37 58 89
Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.
How many starting numbers below ten million will arrive at 89?
题目大意:
通过将一个数各位的平方不断相加,直到遇到已经出现过的数字,可以形成一个数字链。
例如:
44 32 13 10 1 1 85 89 145 42 20 4 16 37 58 89
因此任何到达1或89的数字链都会陷入无限循环。令人惊奇的是,以任何数字开始,最终都会到达1或89。
以一千万以下的数字n开始,有多少个n会到达89?
算法一:常规方法,从2~10000000逐个判断,同时统计结果
#include<stdio.h> #define N 10000000 int fun(int n)
{
int t, sum;
sum = ;
while(n) {
t = n % ;
sum += t * t;
n /= ;
}
return sum;
} void solve(void)
{
int i, sum, t;
sum = ;
for(i = ; i < N; i++) {
t = fun(i);
while() {
if(t == ) {
sum++;
break;
} else if(t == ) {
break;
} else {
t = fun(t);
}
}
}
printf("%d\n",sum);
} int main(void)
{
solve();
return ;
}
算法二(优化):使用一个bool型数组,保存每次结果,由于最大的中间数为9999999产生的:9^2*7 = 567,所以bool型数组的大小开到600足够
#include <stdio.h>
#include <stdbool.h> #define N 10000000 bool a[] = {false}; int fun(int n)
{
int t, sum;
sum = ;
while(n) {
t = n % ;
sum += t * t;
n /= ;
}
return sum;
} void solve(void)
{
int i, sum, t, temp;
sum = ;
for(i = ; i < N; i++) {
t = fun(i);
temp = t;
if(a[temp]) {
sum++;
} else {
while() {
t = fun(t);
if(a[t] || t == ) {
a[temp] = true;
sum++;
break;
} else if(t == ) {
break;
} else {
}
}
}
}
printf("%d\n",sum);
} int main(void)
{
solve();
return ;
}
Answer:
|
8581146 |
(Problem 92)Square digit chains的更多相关文章
- Project Euler 92:Square digit chains C++
A number chain is created by continuously adding the square of the digits in a number to form a new ...
- Project Euler 92:Square digit chains 平方数字链
题目 Square digit chains A number chain is created by continuously adding the square of the digits in ...
- (Problem 74)Digit factorial chains
The number 145 is well known for the property that the sum of the factorial of its digits is equal t ...
- (Problem 34)Digit factorials
145 is a curious number, as 1! + 4! + 5! = 1 + 24 + 120 = 145. Find the sum of all numbers which are ...
- (Problem 33)Digit canceling fractions
The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplif ...
- Project Euler:Problem 63 Powerful digit counts
The 5-digit number, 16807=75, is also a fifth power. Similarly, the 9-digit number, 134217728=89, is ...
- Porject Euler Problem 6-Sum square difference
我的做法就是暴力,1+...+n 用前n项和公式就行 1^2+2^2+....+n^2就暴力了 做完后在讨论版发现两个有趣的东西. 一个是 (1+2+3+...+n)^2=(1^3)+(2^3)+(3 ...
- lintcode:快乐数
快乐数 写一个算法来判断一个数是不是"快乐数". 一个数是不是快乐是这么定义的:对于一个正整数,每一次将该数替换为他每个位置上的数字的平方和,然后重复这个过程直到这个数变为1,或是 ...
- UVA - 10162 Last Digit
Description Problem B.Last Digit Background Give you a integer number N (1<=n<=2*10100). Ple ...
随机推荐
- php7 install memcached extension
#download source code package from git $ git clone https://github.com/php-memcached-dev/php-memcache ...
- jedis入门一
一.下载Jedis的依赖包jedis-2.1.0.jar,然后将其添加到classpath下面. 1. 定义连接:Redis暂时不要设置登录密码 Jedis jedis = new Jedis(&qu ...
- getchar()与EOF
大师级经典的著作,要字斟句酌的去读,去理解.以前在看K&R的The C Programming Language(Second Edition)中第1.5节的字符输入/输出,很迷惑getcha ...
- 3.java.lang.ClassNotFoundException
指定的类不存在 这里主要考虑一下类的名称和路径是否正确即可,通常都是程序试图通过字符串来加载某个类时可能引发 异常 比如: 调用Class.forName(); 或者调用ClassLoad的finaS ...
- SSH Session Recorder
If you want to record your root ssh session create a file .bash_profile . and copy below line by l ...
- Linux-手动释放缓存(Buffer、Cache)
/proc是一个虚拟文件系统,我们可以通过对它的读写操作做为与kernel实体间进行通信的一种手段.也就是说可以通过修改/proc中的文件,来对 当前kernel的行为做出调整.那么我们可以通过调整/ ...
- SendMessage的返回值,就是由相应的响应消息函数的返回值(解释的简洁明了)
SendMessage Return Values The return value specifies the result of the message processing and depend ...
- QSqlDatabase的进一步封装(多线程支持+更加简单的操作)——同时支持MySQL, SQL Server和Sqlite
开发背景: 1.直接用QSqlDatabase我觉得太麻烦了: 2.对于某些数据库,多个线程同时使用一个QSqlDatabase的时候会崩溃: 3.这段时间没什么干货放出来觉得浑身不舒服,就想写一个. ...
- iOS 视图跳转
//跳转 - ( void)present:( id )sender { NSLog ( @"the button,is clicked …" ); // 创建准备跳转的 UIVi ...
- poj2328---"right on"进入下一个case的模板(while)
#include <stdio.h> #include <stdlib.h> #include<string.h> int main() { ]; ,end=; w ...