题目链接:
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=1532
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
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
/* * 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; }
—— 生命的意义,在于赋予它意义。
原创 http://blog.csdn.net/shuangde800 , By D_Double (转载请标明)