CodeForces 289B Polo the Penguin and Matrix

来源：互联网发布：static在java中的作用编辑：程序博客网时间：2024/04/28 22:48

觉得和矩阵没什么关系，直接看成一个数列，从中位数开始，小的增加d，大的减小d，求操作多少次即可。

如果取余不相等，那肯定不可能转换成同一个数字。

#include<stdio.h>#include<iostream>#include<math.h>#include<string.h>#include<iomanip>#include<stdlib.h>#include<ctype.h>#include<algorithm>#include<deque>#include<functional>#include<iterator>#include<vector>#include<list>#include<map>#include<queue>#include<set>#include<stack>#define CPY(A, B) memcpy(A, B, sizeof(A))typedef long long LL;typedef unsigned long long uLL;const int MOD = 1e9 + 7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const double EPS = 1e-9;const double OO = 1e20;const double PI = acos (-1.0);const int dx[] = {-1, 0, 1, 0};const int dy[] = {0, 1, 0, -1};using namespace std;int M[1000010];int main() {    int n,m,d,ans=0,i;    scanf ("%d%d%d",&n,&m,&d);    int num=n*m;    for (i=0; i<num; i++) {        scanf ("%d",&M[i]);        if (M[i]%d!=M[0]%d) {printf ("-1\n"); return 0;}//can not make all matrix elements equal    }    sort (M,M+num);    int mid=M[num/2];    for (i=0; i<num; i++) {        ans+=abs (M[i]-mid) /d;// the minimum number of moves    }    printf ("%d\n",ans);    return 0;}

0 0