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ACdream原创群赛(16) J

2024-07-11 17:37:38   224人阅读


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

求1111111...111我用到打表减少时间复杂。


AC代码如下：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#define M 1000000007
#define min(a,b) (a<b?a:b)
using namespace std;

char a[1005][1005];
long long b[1005];

int main()
{
    int n;
    long long i,j;
    for(i=1;i<=1005;i++)
        b[i]=0;
    for(i=1;i<=1005;i++)
        for(j=1;j<=i;j++)
    {
        b[i]=(b[i]*10+1)%M;
    }
    while(~scanf("%d",&n))
    {
        long long sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%s",&a[i]);
            for(j=0;j<n;j++)
                sum=(sum+((((j+1)*((long long)a[i][j]-'0'))*b[n-j])%M))%M;
        }
        for(j=0;j<n;j++)
        {
            for(i=0;i<n;i++)
                sum=(sum+((((i+1)*((long long)a[i][j]-'0'))*b[n-i])%M))%M;
        }
        cout<<sum<<endl;

    }
    return 0;
}



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