Codeforces Round #289 (Div. 2, ACM ICPC Rules) D. Restoring Numbers 构造 数学

D. Restoring Numbers

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya had two arrays consisting of non-negative integers: a of size n and b of size m. Vasya chose a positive integer k and created ann × m matrix v using the following formula:

Vasya wrote down matrix v on a piece of paper and put it in the table.

A year later Vasya was cleaning his table when he found a piece of paper containing an n × m matrix w. He remembered making a matrix one day by the rules given above but he was not sure if he had found the paper with the matrix v from those days. Your task is to find out if the matrix w that you've found could have been obtained by following these rules and if it could, then for what numbersk, a1, a2, ..., an, b1, b2, ..., bm it is possible.

Input

The first line contains integers n and m (1 ≤ n, m ≤ 100), separated by a space — the number of rows and columns in the found matrix, respectively.

The i-th of the following lines contains numbers wi, 1, wi, 2, ..., wi, m (0 ≤ wi, j ≤ 109), separated by spaces — the elements of the i-th row of matrix w.

Output

If the matrix w could not have been obtained in the manner described above, print "NO" (without quotes) in the single line of output.

Otherwise, print four lines.

In the first line print "YES" (without quotes).

In the second line print an integer k (1 ≤ k ≤ 1018). Note that each element of table w should be in range between 0 and k - 1inclusively.

In the third line print n integers a1, a2, ..., an (0 ≤ ai ≤ 1018), separated by spaces.

In the fourth line print m integers b1, b2, ..., bm (0 ≤ bi ≤ 1018), separated by spaces.

Sample test(s)

input

2 31 2 32 3 4

output

YES10000000070 1 1 2 3 

input

2 21 22 0

output

YES30 1 1 2 

input

2 21 22 1

output

NO

Note

By  we denote the remainder of integer division of b by c.

It is guaranteed that if there exists some set of numbers k, a1, ..., an, b1, ..., bm, that you could use to make matrix w, then there also exists a set of numbers that meets the limits 1 ≤ k ≤ 1018, 1 ≤ ai ≤ 1018, 1 ≤ bi ≤ 1018 in the output format. In other words, these upper bounds are introduced only for checking convenience purposes.

题目给出公式aij = (aai + bbj)% g; 给出vij,要求是否存在这样的数列，若存在则求出aai bbj 和g

首先每行每列，通过 (a[i+1][j+1] - a[i+1][j]) - (a[i][j+1] - a[i][j]) =k1g - k2g = k3g ; (a[i+1][j+1] - a[i][j+1]) - (a[i+1][j] - a[i][j])) = k4g - k5g = k6g,这样求

一下gcd就可以得出g了，若g为0，说明没有冲突，直接置为最大值。若a有数比g大的数，则说明求出的g不能满足题意，直接输出no,

否则，说明不冲突，则要构造出a b即可！如何构造a b呢，直接令aa[0] = 0（因为最后都要取模的，每一个数是几并不影响后面的值）,则bb[0] = a[0][0],通过a[i][0] - a[i-1][0]  = a[i-1]%g,即可构造出来，且构造出

来的一定成立，因为，若有一个数不成立，则这个数与上面的数相减后，就不会是g的倍数，这与前面是矛盾的。这样就ok了！

#define N 105#define M 100005#define maxn 205#define MOD 1000000007int n,m;ll a[N][N],aa[N],bb[N];int main(){    while(S2(n,m)!=EOF)    {       FI(n){            FJ(m){                cin>>a[i][j];            }       }       ll g = 0;       FI(n - 1){            FJ(m - 1){                ll g1 = abs((a[i+1][j+1] - a[i+1][j]) - (a[i][j+1] - a[i][j]));                ll g2 = abs((a[i+1][j+1] - a[i][j+1]) - (a[i+1][j] - a[i][j]));                g = gcd(g,gcd(g1,g2));            }       }       if(g == 0) g = MOD;       bool flag = true;       for(int i=0;i<n && flag;i++){            for(int j = 0;j<m && flag;j++){                if(a[i][j] >= g) flag = false;            }       }       if(!flag){printf("NO\n");continue;}       aa[0] = 0;bb[0] = a[0][0];       for(int i=1;i<n;i++){aa[i] = ( aa[i-1] + a[i][0] - a[i-1][0] + g) % g;}       for(int i=1;i<m;i++){bb[i] = ( bb[i-1] + a[0][i] - a[0][i-1] + g) % g;}       cout<<"YES\n"<<g<<endl;       for(int i=0;i<n;i++){cout<<aa[i]<<" ";}cout<<endl;       for(int i=0;i<m;i++){cout<<bb[i]<<" ";}cout<<endl;    }    return 0;}