HDU-P2097:Sky数[水题]

题目

思路

刚开始拿到题目还觉得有字符串处理之类的内容，不过只要稍微思考一下就能想明白这其实是对N进制数实质的考察。加上最近学习的计算机组成原理，很多都用到了数字逻辑电路的知识，计算机专业要学习的内容真的是一环套一环呀。
普通一点的思路就是按照进制数的计算规则算出各位数字的和，然后分别判断10、12和16进制的计算结果相不相同即可。

代码

代码如下:

#include <iostream>
#include <cstdio>

using namespace std;

int cal16(int n)
{
    int sum = 0;
    while(n > 0) {
        sum += n % 16;
        n /= 16;
    }
    return sum;
}
int cal12(int n)
{
    int sum = 0;
    while(n > 0) {
        sum += n % 12;
        n /= 12;
    }
    return sum;
}
int main(void)
{
    int n;
    int sum;
    while(~scanf("%d", &n)) {
        if(n == 0) break;
        int tmp = n;
        sum = 0;
        while(n > 0) {
            sum+= n%10;
            n/=10;
        }
        if( sum == cal16(tmp)  && sum == cal12(tmp))
            printf("%d is a Sky Number.\n", tmp);
        else printf("%d is not a Sky Number.\n", tmp);
    }
    return 0;
}

题目给出的数据范围其实并不大，所以还可以通过二次变成编程来解决问题:

#include <stdio.h>

int ans[] = {
2992,2993,2994,2995,2996,2997,2998,2999,4470,4471,4472,4473,4474,4475,4970,4971,4972,4973,4974,4975,5460,5461,5462,5463,5464,5465,5466,5467,5468,5469,5960,5961,5962,5963,6456,6457,6458,6459,7440,7441,7442,7443,7444,7445,7446,7447,7448,7449,7940,7941,7942,7943,8436,8437,8438,8439,9424,9425,9426,9427,9428,9429,9924,9925,9926,9927,9928,9929,
};
int binarySearch(int n)
{
    int L = 0, R = sizeof(ans)/sizeof(int)-1;
    while(L <= R) {
        int mid = (L+R)/2;
        if(ans[mid] == n) return 1;
        if(ans[mid] > n) R = mid-1;
        else L = mid+1;
    }
    return 0;
}
int main(void)
{
    int n;
    while(scanf("%d", &n)) {
        if(n == 0) break;
        if(binarySearch(n))
            printf("%d is a Sky Number.\n", n);
        else printf("%d is not a Sky Number.\n", n);
    }
    return 0;
}