Codeforces 2B. The least round way

There is a square matrix n × n, consisting of non-negative integer numbers. You should find such a way on it that

starts in the upper left cell of the matrix;

each following cell is to the right or down from the current cell;

the way ends in the bottom right cell.

Moreover, if we multiply together all the numbers along the way, the result should be the least "round". In other words, it should end in the least possible number of zeros.

Input

The first line contains an integer number n (2 ≤ n ≤ 1000), n is the size of the matrix. Then follow n lines containing the matrix elements (non-negative integer numbers not exceeding 109).

Output

In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.

解题报告:

google翻译是什么东西啊？multiply居然是相加.......

然后就比较简单了:多少个\(0\)可以看成有多少个\(10\)，多少个\(10\)可以看成有多少个\(2\)和\(5\)因子，这样就可以分开处理了，实际上就是分别统计最少经过的\(2\)因子的个数和\(5\)因子的个数,然后取Min就行了，DP方程如同方格取数: \(f[i][j]=Min(f[i-1][j],f[i][j-1])+a[i][j]\) a[i][j]为\((i,j)\)这个位置目前处理的因子的个数

但是注意存在0的情况答案最多为1，如果不存在答案为0的路径，那么就直接走经过0的路径即可

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=1005;
int f[2][N][N],a[N][N],n,p[2][N][N],pre[2][N][N];
char s[N<<1];
void work()
{
	int cnt=0,x;bool flag=false;
	int pi,pj;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=n;j++){
			scanf("%d",&a[i][j]);
			x=a[i][j];
			if(!x){
				flag=true,pi=i,pj=j;
				p[0][i][j]++;p[1][i][j]++;
				continue;
			}
			cnt=0;
			while(x%2==0)x>>=1,cnt++;
			p[0][i][j]=cnt;cnt=0;
			while(x%5==0)x/=5,cnt++;
			p[1][i][j]=cnt;
		}
	for(int k=0;k<=1;k++)
		for(int i=1;i<=n;i++)
			for(int j=1;j<=n;j++){
				if(i-1 && j-1)
				{
					if(f[k][i-1][j]<f[k][i][j-1]){
						f[k][i][j]=f[k][i-1][j];
						pre[k][i][j]=1;
					}
					else{
						f[k][i][j]=f[k][i][j-1];
						pre[k][i][j]=2;
					}
				}
				else{
					if(i-1)f[k][i][j]=f[k][i-1][j],pre[k][i][j]=1;
					else if(j-1)f[k][i][j]=f[k][i][j-1],pre[k][i][j]=2;
				}
				f[k][i][j]+=p[k][i][j];
			}
	int ans=Min(f[0][n][n],f[1][n][n]);
	if(flag && ans>1){
		puts("1");
		for(int i=1;i<pi;i++)putchar('D');
		for(int i=1;i<pj;i++)putchar('R');
		for(int i=pi;i<n;i++)putchar('D');
		for(int i=pj;i<n;i++)putchar('R');
		return ;
	}
	printf("%d\n",ans);
	pi=n;pj=n;int k=(ans==f[0][n][n]?0:1);
	for(int i=1;i<=n+n-2;i++){
		if(pre[k][pi][pj]==1)s[i]='D',pi--;
		else if(pre[k][pi][pj]==2)s[i]='R',pj--;
	}
	for(int i=n+n-2;i>=1;i--)putchar(s[i]);
}

int main()
{
	work();
	return 0;
}