UVA10591 Happy Number【數學】
Let the sum of the square of the digits of a positive integer S0 be represented by S1. In a similar way, let the sum of the squares of the digits of S1 be represented by S2 and so on. If Si = 1 for some i ≥ 1, then the original integer S0 is said to be Happy number. A number, which is not happy, is called Unhappy number. For example 7 is a Happy number since 7 → 49 → 97 → 130 → 10 → 1 and 4 is an Unhappy number since 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4.
Input
The input consists of several test cases, the number of which you are given in the first line of the input. Each test case consists of one line containing a single positive integer N smaller than 10^9.
Output
For each test case, you must print one of the following messages:
Case #p: N is a Happy number.
Case #p: N is an Unhappy number.
Here p stands for the case number (starting from 1). You should print the first message if the number N is a happy number. Otherwise, print the second line.
Sample Input
3
7
4
13
Sample Output
Case #1: 7 is a Happy number.
Case #2: 4 is an Unhappy number.
Case #3: 13 is a Happy number.
問題簡述:(略)
問題分析: 某一個正整數n,對其各位數字分別平方再求和得到一個新數,重複同樣的計算,最終和變成1,則稱n為快樂數;如果出現迴圈變不成1則不是快樂數。
程式說明:
使用set來判重複是一個好做法。
函式ishn()是CV來的,其中包含統計步數的邏輯,參見參考連結。
題記:(略)
AC的C++語言程式如下:
/* UVA10591 Happy Number */
#include <bits/stdc++.h>
using namespace std;
int ishn(int n) {
set<int> s;
int step = 1;
while (n != 1) {
step++;
int sum = 0;
while (n) {
int d = n % 10;
sum += d * d;
n /= 10;
}
n = sum;
if (s.count(n)) break;
else s.insert(n);
}
return n == 1 ? step : 0;
}
int main()
{
int t, caseno = 0, n;
scanf("%d", &t);
while(t--) {
scanf("%d", &n);
if(ishn(n))
printf("Case #%d: %d is a Happy number.\n", ++caseno, n);
else
printf("Case #%d: %d is an Unhappy number.\n", ++caseno, n);
}
return 0;
}