hdoj function 5608 （数学）

function

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 34    Accepted Submission(s): 7

Problem Description

f(x),which is defined on the natural numbers set N,satisfies the following eqaution

N2−3N+2=∑d|Nf(d)

calulate ∑Ni=1f(i) mod 109+7.

Input

the first line contains a positive integer T,means the number of the test cases.

next T lines there is a number N


T≤500,N≤109

only 5 test cases has N>106.

Output

Tlines,each line contains a number,means the answer to the i-th test case.

Sample Input

1 3

Sample Output

2

$1^2-3*1+2=f(1)=0$
$2^2-3*2+2=f(2)+f(1)=0->f(2)=0$
$3^2-3*3+2=f(3)+f(1)=2->f(3)=2$
$f(1)+f(2)+f(3)=2$

//WA的代码：

可能是思路错了

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define ll long long
using namespace std;
const int M=1000000007;
int main()
{
	int t;
	ll sum,num;
	ll n,n1,n2;
	scanf("%d",&t);
	while(t--)
	{
		sum=0;
		scanf("%lld",&n);
		n1=(ll)(n*(n-1))%M;
		sum=(ll)(n1*(n-2))%M;
		sum=(ll)sum/3;
		printf("%lld\n",sum);
	}
	return 0;
}