QAQ的矩阵价值

#include<cstdio>
#include<iostream>
#include<cmath>
#define mod (int)(1e9+7)
using namespace std;
int main()
{
    int t,n;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        long long ans1 = (1+n)*n/2,ans2 = 1,ans3 = (1+n)*n/2;
        long long a1,a2,a3;
        int i,j;
        for(i = 1;i < n;i++)
        {
            a1 = n*(n-i)%mod;
            for(j = n-i-1;j > 0;j--)
                a1 -= j;
            a1 = a1*(int)pow(10,i)%mod;
            ans1 = (ans1+a1)%mod;
        }
        for(i = 1;i < n;i++)
        {
            a2 = (n-i)*(i+1)%mod;
            int k = 1;
            for(j = i+1;j > 0;j--)
            {
                if((n-i-k)==0)
                    break;
                a2 = (a2-(n-i-k))%mod;
                k++;
            }
            a2 = a2*(int)pow(10,i)%mod;
            ans2 = (ans2+a2)%mod;
        }
        for(i = 1;i < n;i++)
        {
            a3 = (1+n-i)*(n-i)/2%mod;
            a3 = a3*(int)pow(10,i)%mod;
            ans3 = (ans3+a3)%mod;
        }
        printf("%lld %lld %lld\n",ans1,ans2,ans3);
    }
    return 0;
}

 

先记录下。。因为还不会逆元。大数无法处理！！！  注意：这个不是AC代码