ACdream 1139(Sum-逆元)

J - 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

3123456789

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;}