HDU 2608 0 or 1

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0 or 1

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3837    Accepted Submission(s): 1095

Problem Description

Solving problem is a interesting thing. Yifenfei like to slove different problem,because he think it is a way let him more intelligent. But as we know,yifenfei is weak in math. When he come up against a difficult math problem, he always try to get a hand. Now the problem is coming! Let we
define T(n) as the sum of all numbers which are positive integers can divied n. and S(n) = T(1) + T(2) + T(3)…..+T(n).

 

Input

The first line of the input contains an integer T which means the number of test cases. Then T lines follow, each line consists of only one positive integers n. You may assume the integer will not exceed 2^31.

 

Output

For each test case, you should output one lines of one integer S(n) %2. So you may see the answer is always 0 or 1 .

 

Sample Input

3123 

 

Sample Output

100
Hint
Hint S(3) = T(1) + T(2) +T(3) = 1 + (1+2) + (1+3) = 8     S(3) % 2 = 0 
 

 

Author

yifenfei

 

Source

奋斗的年代

 

T(n)是所有能够被n整除的数的和

对于这题，我们可以先打个表。找一下规律。

#include<iostream>#include<cstdio>#define N 200int T[N],S[N];int main(){    int t,n,i;    for(n=1; n<N; n++)    {        for(i=1; i<=n; i++)        {            if(n%i==0)            {                T[n] += i;            }        }        S[n] = S[n-1]+T[n];        printf("n=%d,T[n]=%d,S[n]=%d\n",n,T[n],S[n]);    }}

打表的结果

我们可以得出结论，当n为完全平方数或者是完全平方数的两倍时,T[n]便为奇数

如果S【n】中有奇数个   奇数T[n]  那么S【n】便为奇数，反之便为偶数。

那么这题就转换为：找T【1】~T【n】中有几个奇数

也就转换为

找1~n中有几个完全平方数，和有几个是完全平方数二倍的数

1~n中完全平方数的个数为  sqrt(n)（取整）

1~n中是完全平方数二倍的数的个数为   sqrt(n/2)(取整)

那么这题的代码就很简单了

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<cmath>using namespace std;int main(){    int t;    cin>>t;    while(t--)    {        int n;        scanf("%d",&n);        int sum1=(int)sqrt(n*1.0);///计算完全平方数        int sum2=(int)sqrt((n*1.0)/2.0);///计算完全平方数的二倍的个数        printf("%d\n",(sum1+sum2)%2);    }    return 0;}