215 = 32768 and the sum of its digits is 3 + 2 + 7 + 6 + 8 = 26.

What is the sum of the digits of the number 21000?

题目大意:

题目大意:

215 = 32768 并且其各位之和为 is 3 + 2 + 7 + 6 + 8 = 26.

21000 的各位数之和是多少?

// (Problem 16)Power digit sum
// Completed on Sun, 17 Nov 2013, 15:23
// Language: C
//
// 版权所有(C)acutus (mail: acutus@126.com)
// 博客地址:http://www.cnblogs.com/acutus/
#include <stdio.h>
#include <stdbool.h> void solve(void)
{
int a[] = {};
int n, sum, i, j;
n = sum = ;
a[] = ;
for(i = ; i < ; i++) { //以1000进制的方法存储
for(j = ; j <= n; j++) {
a[j] *= ;
}
for(j = ; j <= n; j++) {
if(a[j] >= ) {
a[j] %= ;
a[j+]++;
n++;
}
}
}
for(i = ; i <= n; i++) {
sum += a[i] / ;
a[i] %= ;
sum += a[i] / ;
a[i] %= ;
sum += a[i] / ;
a[i] %= ;
sum += a[i] / ;
a[i] %= ;
sum += a[i]; }
printf("%d\n",sum);
} int main(void)
{
solve();
return ;
}
Answer:
1366

(Problem 16)Power digit sum的更多相关文章

  1. project euler 16:Power digit sum

    >>> sum([int(i) for i in str(2**1000)]) 1366 >>>

  2. (Problem 17)Number letter counts

    If the numbers 1 to 5 are written out in words: one, two, three, four, five, then there are 3 + 3 + ...

  3. (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 ...

  4. (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 ...

  5. (Problem 33)Digit canceling fractions

    The fraction 49/98 is a curious fraction, as an inexperienced mathematician in attempting to simplif ...

  6. (Problem 13)Large sum

    Work out the first ten digits of the sum of the following one-hundred 50-digit numbers. 371072875339 ...

  7. (Problem 46)Goldbach's other conjecture

    It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a ...

  8. (Problem 29)Distinct powers

    Consider all integer combinations ofabfor 2a5 and 2b5: 22=4, 23=8, 24=16, 25=32 32=9, 33=27, 34=81, ...

  9. (Problem 28)Number spiral diagonals

    Starting with the number 1 and moving to the right in a clockwise direction a 5 by 5 spiral is forme ...

随机推荐

  1. poj 2135 Farm Tour 费用流

    题目链接 给一个图, N个点, m条边, 每条边有权值, 从1走到n, 然后从n走到1, 一条路不能走两次,求最短路径. 如果(u, v)之间有边, 那么加边(u, v, 1, val), (v, u ...

  2. nginx启动、开机自启动、重启、关闭

    yum -y  install nginx # yum info nginx Loaded plugins: fastestmirror Loading mirror speeds from cach ...

  3. 超轻量级PHP SQL数据库框架

    <?php /** * ! Medoo 0.8.5 - Copyright 2013, Angel Lai - MIT license - http://medoo.in */ class me ...

  4. IPv4头部结构具体解释

    IPv4头部结构具体解释 下面为书中原文摘录: $(function () { $('pre.prettyprint code').each(function () { var lines = $(t ...

  5. jsp 声明类的使用

    能够在"<%!"和"%>"之间声明一个类,该类在JSP页面内有效,即在JSP页面的Java程序片部分能够使用该类创建对象.在以下的样例中,我们定义了 ...

  6. python下yield(生成器)

    python下的协程: #encoding=utf-8 """ 协程----微小的进程 yield生成器-----生成一个可迭代对象比如list, tuple,dir 1 ...

  7. js动态创建表格方法

    window.onload = function(){ var table = document.createElement('table'); table.border = 1; table.wid ...

  8. CSS的position(位置)

    position: 位置,absolute绝对位置,相对于浏览器边界的位置:relative相对位置,相对于它本应该出现的位置.fixed:固定位置,它不会随着滚动. 设置好position之后,就可 ...

  9. Main function

    Main function A program shall contain a global function named main, which is the designated start of ...

  10. python成长之路10——socketserver源码分析

    s = socket.socket(socket.AF_INET,socket.SOCK_STREAM,0) 参数一:地址簇 socket.AF_INET ipv4(默认) socket.AF_INE ...