首页 > ACM题库 > HDU-杭电 > HDU 1197 Specialized Four-Digit Numbers[解题报告] C++
2013
12-04

HDU 1197 Specialized Four-Digit Numbers[解题报告] C++

Specialized Four-Digit Numbers

问题描述 :

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 = 1*1728 + 8*144 + 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.)

输入:

There is no input for this problem.

输出:

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.

样例输入:

There is no input for this problem.

样例输出:

2992
2993
2994
2995
2996
2997
2998
2999

题目大意:从2292开始,查找以10进制表示时,各数位上数字之和等于以12、16进制形式表示时的各数位上数字之和

解题思路:依据题意写就行。。。。。

代码如下:

/*
 * 1197_1.cpp
 *
 *  Created on: 2013年8月10日
 *      Author: Administrator
 */

#include <stdio.h>


int mod(int n , int b){
	int sum = 0;
	while(n){
		sum += n%b;
		n /= b;
	}
	return sum;
}
int main(){
	int i ;
	for(i = 2992 ; i <= 9999 ; ++i){
		if(mod(i,10) == mod(i,12) &&mod(i,10) == mod(i,16)){
			printf("%d\n",i);
		}
	}
}


  1. 第一句可以忽略不计了吧。从第二句开始分析,说明这个花色下的所有牌都会在其它里面出现,那么还剩下♠️和♦️。第三句,可以排除2和7,因为在两种花色里有。现在是第四句,因为♠️还剩下多个,只有是♦️B才能知道答案。