UVa 10591 - Happy Number

类型：哈希表

原题：
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.

样例输入：
3
7
4
13

样例输出：
Case #1: 7 is a Happy number.
Case #2: 4 is an Unhappy number.
Case #3: 13 is a Happy number.

题目大意：
所谓的Happy数字，就是给一个正数s，然后计算它每一个位上的平方和，得到它的下一个数，然后下一个数继续及选每位上的平方和……如果一直算下去，没有出现过之前有出现过的数字而出现了1，那么恭喜，这就是个Happy Number.
如果算下去的过程中出现了一个之前出现过的，那么就不Happy了。

思路与总结：
这题算是最简单的一种hash应用，学术一点的说法就是直接寻址表
这题最大的数是10^9, 那么各个位数上都最大的话就是9个9，平方和为9*9*9 = 729, 所以只要开个730+的数组即可。
对于出现过的数字，就在相应的数组下标那个元素标志为true

[cpp]
/*
* UVa 10591 - Happy Number
* Time: 0.008s (UVa)
* Author: D_Double
*/
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int hash[810];

inline int getSum(int n){
    int sum=0;
    while(n){
        int t = n%10;
        sum += t*t;
        n /= 10;
    }
    return sum;
}

int main(){
    int T, cas=1;
    scanf("%d", &T);
    while(T--){
        int N, M;
        memset(hash, 0, sizeof(hash));
        scanf("%d", &N);
        M = N;
        bool flag=false;
        int cnt=1;
        while(M=getSum(M)){
            if(M==1) {flag=true; break;}
            else if(hash[M] || M==N){break;}
            hash[M] = 1;
        }

        printf("Case #%d: ", cas++);
        if(flag) printf("%d is a Happy number.\n", N);
        else printf("%d is an Unhappy number.\n", N);
    }
    return 0;
}

作者：shuangde800

补充：软件开发 , C++ ,