浙大保研2019年上机题 7-1 Happy Numbers (20分)

标签：个数   for   输出   stay   integer   ++   equals   turn   from   

7-1 Happy Numbers (20分)

A happy number is defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits in base-ten, and repeat the process until the number either equals 1 (where it will stay), or it loops endlessly in a cycle that does not include 1. Those numbers for which this process ends in 1 are happy numbers and the number of iterations is called the degree of happiness, while those that do not end in 1 are unhappy numbers (or sad numbers). (Quoted from Wikipedia)

For example, 19 is happy since we obtain 82 after the first iteration, 68 after the second iteration, 100 after the third iteration, and finally 1. Hence the degree of happiness of 19 is 4.

On the other hand, 29 is sad since we obtain 85, 89, 145, 42, 20, 4, 16, 37, 58, and back to 89, then fall into an endless loop. In this case, 89 is the first loop number for 29.

Now your job is to tell if any given number is happy or not.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (≤100). Then N lines follow, each contains a positive integer (no more than 104) to be tested.

Output Specification:

For each given number, output in a line its degree of happiness if it is happy, or the first loop number if it is sad.

Sample Input:

3
19
29
1

Sample Output:

4
89
0

happy number指的是，一个数字，各位数字的平方进行相加，可以进行迭代，迭代一次，幸福度+1， 如果造成循环，则输出循环的数字，如果不造成循环（最终为1），则输出幸福度。

#include <iostream>
#include <cmath>
#include <unordered_map>
using namespace std;
int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    int N, tmp;
    scanf("%d", &N);
    while(N--) {
        scanf("%d", &tmp);
        unordered_map<int, bool> m;
        m[tmp] = 1;
        int happy = 0;
        while(tmp != 1) {
            string s = to_string(tmp);
            tmp = 0;
            for(int i = 0; i < s.size(); i++) 
                tmp += pow((s[i] - ‘0‘), 2);
            if(m[tmp] == 1) break;
            else m[tmp] = 1;
            happy++;
        }
        printf("%d\n", tmp == 1 ? happy: tmp);
    }
    return 0;
}

浙大保研2019年上机题 7-1 Happy Numbers (20分)

标签：个数   for   输出   stay   integer   ++   equals   turn   from   

原文地址：https://www.cnblogs.com/littlepage/p/13194679.html