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最新推荐文章于 2022-10-24 16:06:31 发布

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最新推荐文章于 2022-10-24 16:06:31 发布

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Another Sum Problem

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1472 Accepted Submission(s): 407

Problem Description

FunnyAC likes mathematics very much. He thinks mathematics is very funny and beautiful.When he solved a math problem he would be very happy just like getting accepted in ACM.Recently, he find a very strange problem.Everyone know that the sum of sequence from 1 to n is n*(n + 1)/2. But now if we create a sequence which consists of the sum of sequence from 1 to n. The new sequence is 1, 1+ 2, 1+2+3, .... 1+2+...+n. Now the problem is that what is the sum of the sequence from1 to 1+2+...+n .Is it very simple? I think you can solve it. Good luck!

Input

The first line contain an integer T .Then T cases followed. Each case contain an integer n (1 <= n <= 10000000).

Output

For each case,output the sum of first n items in the new sequence. Because the sum is very larger, so output sum % 20090524.

Sample Input

Sample Output

Source

HDU 2009-5 Programming Contest

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求证：1*2/2+2*3/2+3*4/2+……+n(n+1)/2=n(n+1)(n+2)/6

证：1）n=1时，左边=1*2/2=1,右边=1*2*3/6=1,所以n=1时等式成立。
2）设n=k时等式1*2/2+2*3/2+3*4/2+……+k(k+1)/k=k(k+1)(k+2)/6 成立。则n=k+1时
左边=1*2/2+2*3/2+……+k(k+1)/2+(k+1)(k+2)/2
=k(k+1)(k+2)/6+(k+1)(k+2)/2
=(k+1)(k+2)(k+3)/6
=(k+1)[(k+1)+1]*[(k+1)+2]/6.就是说n=k+1时等式成立。

因此对一切正整数n，等式都成立。

附代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
__int64 m,n,k;
int main()
{
	scanf("%I64d",&k);
	while(k--)
	{
		scanf("%I64d",&n);//公式并不难推 
		m=n*(n+1)/2%(20090524*3);//但是一乘就越界 
		m=m*(n+2)/3%20090524;//迫不得已 
		printf("%I64d\n",m);//就ac了 
	}
}