题目传送门：https://codeforces.com/problemset/problem/2/B
题目大意：
给定 (n imes n) 的数表，求一条从左上角到右下角的路径，使得路径上各数相乘所得的后置零尽可能少，输出操作序列（DR）
由于后置零仅可能由(2 imes 5)得到，故我们将所有数中的2,5质因子个数求出来，然后找到一条2的个数或5的个数最少的路径即可。
/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
static char buf[1000000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
int f=1; char ch=gc();
for (;ch<'0'||ch>'9';ch=gc()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=gc()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
template<typename T>inline T read(T x){
int f=1; char ch=getchar();
for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
for (;ch>='0'&&ch<='9';ch=getchar()) x=(x<<1)+(x<<3)+ch-'0';
return x*f;
}
inline void print(int x){
if (x<0) putchar('-'),x=-x;
if (x>9) print(x/10);
putchar(x%10+'0');
}
const int N=1e3;
int F[2][N+10][N+10],opt;
pii Frm[2][N+10][N+10];
char Ans[N+10];
char Direction(int sx,int sy,int ex,int ey){
if (sx+1==ex) return 'D';
if (sy+1==ey) return 'R';
return 0;
}
void solve(int x,int y){
int len=0;
while (x>1||y>1){
pii temp=Frm[opt][x][y];
Ans[len++]=Direction(temp.Fi,temp.Se,x,y);
x=temp.Fi,y=temp.Se;
}
reverse(Ans,Ans+len);
printf("%s
",Ans);
}
int main(){
// freopen(".in","r",stdin);
// freopen(".out","w",stdout);
int n=read(0),Flag=0,Zx,Zy;
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
int x=read(0);
if (!x){
Zx=i,Zy=j;
Flag|=1;
continue;
}
while (x%2==0) F[0][i][j]++,x/=2;
while (x%5==0) F[1][i][j]++,x/=5;
}
}
for (int K=0;K<2;K++){
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
if (i==1&&j==1) continue;
if (i==1){
F[K][i][j]+=F[K][i][j-1];
Frm[K][i][j]=MK(i,j-1);
continue;
}
if (j==1){
F[K][i][j]+=F[K][i-1][j];
Frm[K][i][j]=MK(i-1,j);
continue;
}
F[K][i][j]+=min(F[K][i-1][j],F[K][i][j-1]);
Frm[K][i][j]=F[K][i-1][j]<F[K][i][j-1]?MK(i-1,j):MK(i,j-1);
}
}
}
opt=F[0][n][n]<F[1][n][n]?0:1;
int Min=min(F[0][n][n],F[1][n][n]);
if (Min>1&&Flag){
printf("1
");
for (int i=1;i<Zx;i++) putchar('D');
for (int i=1;i<Zy;i++) putchar('R');
for (int i=Zx;i<n;i++) putchar('D');
for (int i=Zy;i<n;i++) putchar('R');
return 0;
}
printf("%d
",Min),solve(n,n);
return 0;
}