杭电OJ | 1197 A hard puzzle 2031 进制转换

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本文探讨了寻找特定四位数的挑战，这些数在十进制、十六进制和十二进制下的数字之和相等。通过算法实现，列举了满足条件的所有四位数，展示了进制转换与数学运算的巧妙结合。

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1197 A hard puzzle

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.

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 Output

2992

2993

......

要点：进制转换问题

#include <stdio.h>
#include <math.h>
int sum(int start, int base);
int main(){
	int start;
	for(start = 2991; start<=9999; start++){
		int dec = sum(start,10);
		int duo = sum(start,12);
		int hex = sum(start,16);
		if(duo==dec&&hex==dec){
			printf("%d\n",start);
		}	
	}
	return 0;
}
int sum(int start, int base){
	int i,remain;
	int sum = 0;
	for(i=0;start>0;i++){
		remain = start%base;
		sum = sum+remain;
		start = start/base;
	}
	return sum;	
}

2031 进制转换

Problem Description 输入一个十进制数N，将它转换成R进制数输出。

Input 输入数据包含多个测试实例，每个测试实例包含两个整数N(32位整数)和R（2<=R<=16, R<>10）。

Output 为每个测试实例输出转换后的数，每个输出占一行。如果R大于10，则对应的数字规则参考16进制（如10用A表示）。

Sample Input

7 2

23 12

-4 3

Sample Output

111

-11

#include <stdio.h>
int main() {
    int num, base;
    while (scanf("%d %d", &num, &base) != EOF) {
        int quo = num;
        int flag = 1;
        if (quo < 0) {
            quo = -quo;
            flag = 0;
        }
        int remain, j;
        int i = 0;
        char output[50];
        while (quo != 0) {
            remain = quo % base;
            if (remain > 9) {
                if (remain == 10) {
                    output[i] = 'A';
                } else if (remain == 11) {
                    output[i] = 'B';
                } else if (remain == 12) {
                    output[i] = 'C';
                } else if (remain == 13) {
                    output[i] = 'D';
                } else if (remain == 14) {
                    output[i] = 'E';
                } else if (remain == 15) {
                    output[i] = 'F';
                }
            } else {
                output[i] = remain + '0';
            }
            quo = quo / base;
            i++;
        }
        if (flag == 0) {
            printf("-");
        }
        for (j = i - 1; j >= 0; j--) {
            printf("%c", output[j]);
        }
        printf("\n");
    }
    return 0;
}