ZOJ2405 Specialized Four-Digit Numbers 2017-04-18 20:43 44人阅读 评论(0) 收藏

Specialized Four-Digit Numbers

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 189312, 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

2992

2993

2994

2995

2996

2997

2998

2999

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Source: Pacific Northwest 2004

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题目的意思是找出所有4位整数，要求他的10进制，12进制，16进制各位和相等

思路：暴力枚举

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <set>
#include <stack>
#include <map>
#include <functional>
#include <bitset>
#include <string>

using namespace std;

#define LL long long
#define INF 0x3f3f3f3f

int main()
{
    for(int i=1000;i<10000;i++)
    {
        int x1=i;
        int ans1=0;
        while(x1)
        {
            ans1+=x1%10;
            x1/=10;
        }
        int x2=i;
        int ans2=0;
        while(x2)
        {
            ans2+=x2%12;
            x2/=12;
        }
        int x3=i;
        int ans3=0;
        while(x3)
        {
            ans3+=x3%16;
            x3/=16;
        }
        if(ans1==ans2&&ans1==ans3)
            printf("%d\n",i);

    }
    return 0;
}