UVA - 10591
| Time Limit: 3000MS | Memory Limit: Unknown | 64bit IO Format: %lld & %llu |
Description
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Problem C |
Happy Number |
|
Time Limit |
1 Second |
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 i3 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 109.
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.
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Sample Input |
Output for Sample Input |
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3 7 4 13 |
Case #1: 7 is a Happy number. Case #2: 4 is an Unhappy number. Case #3: 13 is a Happy number. |
Problemsetter: Mohammed Shamsul Alam
International Islamic University Chittagong
Special thanks to Muhammad Abul Hasan
Source
思路：简单题，直接模拟过去即可，这里用set记录某个数出现没，出现了就可以判断出有环，即不是Happy Number，刚开始没想清循环结束条件，结果TLE了一次
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
int main() {
int N, cas = 1;
scanf("%d", &N);
while(N--) {
int a, flag = 0;
scanf("%d", &a);
set<int> num;
num.insert(a);
int t = a;
while(1) {
if(t == 1) {
flag = 1;
break;
}
int tmp = t;
t = 0;
while(tmp) {
int b = tmp % 10;
t += b * b;
tmp /= 10;
}
if(num.count(t)) break;
else num.insert(t);
}
if(flag) {
printf("Case #%d: %d is a Happy number.\n", cas++, a);
}
else printf("Case #%d: %d is an Unhappy number.\n", cas++, a);
}
return 0;
}
UVA - 10591 - Happy Number （STL）
原文：http://blog.csdn.net/u014355480/article/details/44540957