J - Sum
Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)
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 须要累加的值为 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; }