(hdu step 1.2.8)Specialized Four-Digit Numbers(求一个数字各个数位上数字的和)-CFANZ编程社区

题目：

Specialized Four-Digit Numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2027 Accepted Submission(s): 1349

Problem Description Find and list all four-digit numbers in decimal notation that have the property that the sum of its four digits equals the sum of its digits when represented in hexadecimal (base 16) notation and also equals the sum of its digits when represented in duodecimal (base 12) notation. For example, the number 2991 has the sum of (decimal) digits 2+9+9+1 = 21. Since 2991 = 11728 + 8144 + 9*12 + 3, its duodecimal representation is 1893(12), and these digits also sum up to 21. But in hexadecimal 2991 is BAF16, and 11+10+15 = 36, so 2991 should be rejected by your program. The next number (2992), however, has digits that sum to 22 in all three representations (including BB016), so 2992 should be on the listed output. (We don't want decimal numbers with fewer than four digits - excluding leading zeroes - so that 2992 is the first correct answer.)
Input There is no input for this problem.
Output Your output is to be 2992 and all larger four-digit numbers that satisfy the requirements (in strictly increasing order), each on a separate line with no leading or trailing blanks, ending with a new-line character. There are to be no blank lines in the output. The first few lines of the output are shown below.
Sample Input There is no input for this problem.
Sample Output 29922993299429952996299729982999

Source Pacific Northwest 2004
Recommend Ignatius.L

题目大意：

列出10进制、12进制、16进制表示时，各个数位上的数字之和相等的所有4位数。

题目分析：

求一个数字的各个数位上的数字之和。

代码如下：

/*
 * i.cpp
 *
 *  Created on: 2015年1月28日
 *      Author: Administrator
 */

#include <iostream>
#include <cstdio>
#include <cstdlib>


using namespace std;


/**
 * 求一个数字用base进制表示时各个数位上的数字的值
 * num:十进制表示的数字
 * base: 进制
 *
 */
int getSum(int num,int base){
	char arr[10];
	itoa(num,arr,base);//需要注意的是itoa()这个函数返回的大于10的书用a表示

//	cout << arr << endl;
	int sum = 0;
	int i=0;
	while(arr[i] != '\0'){//如果这个字符串还没有到末尾
		if(arr[i] >= 'a'){//如果这是一个大浴室的数
			sum += arr[i]-'a' + 10;//后面别忘记+10
		}else{//如果这是一个小于10的数
			sum += arr[i]-'0';
		}
		i++;
	}

	return sum;
}


int main(){
	int i;
	for(i = 1000 ; i < 10000 ; ++i){
		if( (getSum(i,10) == getSum(i,12)) && (getSum(i,10) == getSum(i,16))){
			printf("%d\n",i);
		}
	}


//	printf("%d\n",getSum(2992,10));
//	printf("%d\n",getSum(2992,12));
//	printf("%d\n",getSum(2992,16));

	return 0;
}