ACdream： Sum

Sum

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

SubmitStatisticNext
Problem

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

题目主要是导出公式：

如n行n列的每一行的和sum=1111.....111（n个1）*A1+111...111（n-1个1）*2*A2+.........+11*(n-1)*An-1+1*n*An;

好吧。。这么写果然还是不怎么完整。。如今补充一下。。

拿第一行 1 2 3来说。。

1 仅仅能由他本身， 即 sum+=A1。  2 能有2， 12。则 sum+=A2+10*A1+A2;   3 能有3， 23 ，123，则：

sum+=A3+10*A2+A3+100*A1+10*A2+A3......合算的sum=111*A1+11*2*A2+1*3*A3;也就是上面说的那个公式。

别忘了计算列的sum.

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>

#define M 1000000007
#define f1(i, n)  for(int i=1; i<=n; i++)
#define f2(i, n)  for(int i=0; i<n; i++)
#define f3(j, n)  for(int j=0; j<n; j++)　

using namespace std;

char a[1050][1050];
long long int b[1050];

int main()
{
    int n;
    int k=1050;
    f1(i, k)
       b[i]=0;
    f1(i, k)
       for(int j=1; j<=i; j++)
        b[i]=(b[i]*10+1)%M;
    while(~scanf("%d",&n))
    {
        k=n;
        long long int sum=0;
        f2(i, k)
        {
            scanf("%s",&a[i]);
           // f3(j, k)
           for(int j=0; j<n; j++)
                sum=(sum+((((j+1)*((long long)a[i][j]-'0'))*b[n-j])%M))%M;
        }
        //f3(j, k)
        for(int j=0; j<n; j++)
        {
            f2(i, k)
                sum=(sum+((((i+1)*((long long)a[i][j]-'0'))*b[n-i])%M))%M;
        }
        cout<<sum<<endl;
    }

    return 0;
}