HDU 4342 History repeat itself

最新推荐文章于 2021-02-04 23:00:44 发布

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本文介绍了一个编程挑战，任务是找出第N个正的非平方数及其相关公式计算结果。问题来源于一位教授对学生的考验，涉及循环和数学运算。

History repeat itself

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 488 Accepted Submission(s): 244

Problem Description

Tom took the Discrete Mathematics course in the 2011,but his bad attendance angered Professor Lee who is in charge of the course. Therefore, Professor Lee decided to let Tom face a hard probability problem, and announced that if he fail to slove the problem there would be no way for Tom to pass the final exam.
As a result , Tom passed.
History repeat itself. You, the bad boy, also angered the Professor Lee when September Ends. You have to faced the problem too.
The problem comes that You must find the N-th positive non-square number M and printed it. And that's for normal bad student, such as Tom. But the real bad student has to calculate the formula below.

So, that you can really understand WHAT A BAD STUDENT YOU ARE!!

Input

There is a number (T)in the first line , tell you the number of test cases below. For the next T lines, there is just one number on the each line which tell you the N of the case.
To simplified the problem , The N will be within 2³¹ and more then 0.

Output

For each test case, print the N-th non square number and the result of the formula.

Sample Input

Sample Output

不知为何用sqrt求那个n的时候就WA。。。哎，了解的不够！

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
int main( )
{
    int cas;
    __int64 sum , n , m , ans;
    scanf("%d",&cas);
    while( cas-- )
    {
           sum = 0;
           n = 1;
           scanf("%I64d",&m);
           while( n * (n+1) < m )
                  n++;
           ans = n+m;
           sum = (ans - n*n + 1) * n ;
           while( --n )
           {
                  sum += n * (2*n+1);       
           }
           printf("%I64d %I64d\n",ans , sum);
    }
    return 0;
}