ACdream 1139(Sum-逆元)

J - Sum

Time Limit:  2000/1000MS (Java/Others)  Memory Limit:  128000/64000KB (Java/Others)

Submit Status

Problem Description

You are given an N*N digit matrix and you can get several horizontal or vertical digit strings from any position.

For example:

123

456

789

In first row, you can get 6 digit strings totally, which are 1,2,3,12,23,123.

In first column, you can get 6 digit strings totally, which are 1,4,7,14,47,147.

We want to get all digit strings from each row and column, and write them on a paper. Now I wonder the sum of all number on the paper if we consider a digit string as a complete decimal number.

Input

The first line contains an integer N. (1 <= N <= 1000)

In the next N lines each line contains a string with N digit.

Output

Output the answer after module 1,000,000,007(1e9+7)。

Sample Input

3
123
456
789

Sample Output

2784

本题暴力会T

所以简化公式

对于同行/列 需要累加的值为 a1*111+a2*22+a3*3

发现规律sum=∑a(10^(n-i+1)-1)/9*i %F

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXN (1000+10)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;
int n;
char a[MAXN][MAXN];
ll p10[MAXN]={0};
ll pow2(ll b)  
{
   if (b==1) return 10;  
   if (b==0) return 1;  
   if (p10[b]) return p10[b];
   ll p=pow2(b/2)%F;  
   p=(p*p)%F;  
   if (b&1)  
   {  
       p=(p*10)%F;  
   }  
   p10[b]=p;
   return p;  
}  
ll pow2(ll a,ll b)
{
	if (b==1) return a;
	if (b==0) return 1;
	ll p=pow2(a,b/2)%F;
	p=p*p%F;
	if (b&1)
	{
		p=(p*a)%F;
	}
	return p;
}
ll tot[MAXN]={0};
ll mulinv(ll a)
{
	return pow2(a,F-2);
}
int main()
{
//	freopen("sum.in","r",stdin);
//	freopen("sum.out","w",stdout);
	scanf("%d",&n);
	For(i,n)
	{
		scanf("%s",a[i]+1);
		
	}
	/*
	For(i,n)
	{
		For(j,n) cout<<a[i][j];
		cout<<endl;
	}
	*/
	For(i,n)
	{
		For(j,n) tot[i]+=a[i][j]-'0'+a[j][i]-'0';		
	}
	
	
//	For(i,n) cout<<tot[i]<<endl;
	
//	cout<<mul(pow2(10,1232),mulinv(pow2(10,1232)))<<endl;
//	cout<<mulinv(9);

	ll c9=mulinv(9);
	
	For(i,n) p10[i]=pow2(i);
	
	
	ll ans=0;
	For(i,n)
	{
		ll t=sub(p10[n-i+1],1),a=tot[i];
		t=mul(t,c9);
		t=mul(a,t);
		ans=add(ans,mul(t,i));		
	}
	cout<<ans<<endl;
	
	
	return 0;
}