codechef April challenge （2）（学习去优化提高效率）

Counting Matrices

Problem code: CNPIIM

此题逆向解决问题，先求出乘积为i的数对有多少个，到询问的时候直接输出就行了！

Lira loves Linear Algebra and she is especially keen about matrix :).

Today, she learnt some properties about matrices, namely, she learnt about what the trace of a matrix is, as her teacher gave her many exercises for her to practice.

As we know she is pretty clever, she rapidly came up with some definitions of her own and devised a somewhat harder version of the problem initially proposed by her teacher.

Namely, she defines a Positive Invertible Integer Matrix as being an invertible 2x2 matrix composed only of positive (i.e. greater than 0) integer elements and whose determinant is greater than 0.

Now, she is interested in counting how many such matrices are there, such that their trace is equal toN .

It's your turn to solve Lira's problem :D

Input

The first line of the input contains an integer T denoting the number of test cases. The description ofT test cases follows.

Each test case consist of single integer N, the trace of the matrix.

Output

For each test case, output a single line containing the number of Positive Invertible Integer Matrices such that their trace is equal toN and dterminant is positive.

Constraints:

• 1 ≤ T ≤ 50
• 3 ≤ N ≤ 2500

 

Example

Input:13Output:2

Explanation

The only two matrices that exist with trace equal to 3 and that satisfy all the given conditions are:

#include<stdio.h>#define N 2500const int MAX=N*N/4;int dp[MAX],num[MAX];//计算乘积小于x的数对的个数void unit(){    //计算乘积为i的数对的对数，逆向得到存储在数组里    int i;     for(i=1;i<=MAX;i++)        {            for(int j=1;j<=MAX;j++)            {                if(i*j>MAX) break;                dp[i*j]++;            }        }       for(i=1;i<MAX;i++)        {            num[i]=num[i-1]+dp[i];        }}int main(){   unit();   int a1,t,n;   scanf("%d",&t);   for(;t>0;t--)   {   long long  sum=0;       scanf("%d",&n);       for(a1=1;a1<=(n-1)/2;a1++)       {           int tri=a1*(n-a1);           sum+=num[tri-1];       }       if(n%2!=0)  printf("%lld\n",2*sum);       else   printf("%lld\n",sum*2+num[n/2*n/2-1]);   }   return 0;}