CodeForces - 2B

B. The least round way

time limit per test

2 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

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.

Examples

input

3
1 2 3
4 5 6
7 8 9

output

0
DDRR

 题意：

一个数字矩阵，只能向下走或者向右走，从左上角，到右下角所有数字的乘积末尾最少有几个0，输出其个数，并且输出路径

做法：

预处理，每个数字有多少个2的质因子和5的质因子,分别储存为a[i][j][0]和a[i][j][1]

记忆化搜索,f[x][y][0]为x，y为终点经过的个点的2的质因子的个数的和

f[x][y][1]为x，y为终点经过的个点的5的质因子的个数的和

f[x][y][k] = min(f[x-1][y][k],f[x][y-1][k] )+a[x][y][k]

特殊处理0的情况，把0当成10,去处理，看最小答案是否大于1，如果大于1，那么最优答案就是1，也就是经过0的任意路径。

  1 #include<iostream>
  2 using namespace std;
  3 #include<cstdio>
  4 #include<cstring>
  5 int f[1002][1002][2];
  6 int a[1002][1002][2];
  7 int fx[1002][1002][2];
  8 int v[1002][1002][2];
  9 int dfs(int x,int y,int k){
 10     if(x==1&&y==1){
 11         if(k==0)
 12             return a[1][1][0];
 13         if(k==1)
 14             return a[1][1][1];
 15     }
 16     if(x<=0||y<=0)
 17         return 1e7;
 18     if(v[x][y][k]!=0)
 19         return f[x][y][k];
 20     v[x][y][k]=1;
 21     int ans;
 22     ans = dfs(x-1,y,k);
 23     fx[x][y][k] = 0;
 24     int ans2=dfs(x,y-1,k);
 25     if(ans2<ans){
 26         ans = ans2;
 27         fx[x][y][k] =1;
 28     }
 29     if(k==0)
 30     f[x][y][k] = ans + (a[x][y][k]);
 31     if(k==1)
 32     f[x][y][k] = ans + (a[x][y][k]);
 33     return f[x][y][k];
 34 }
 35 void out(int x,int y,int k){
 36 //    cout<<x<<" "<<y<<endl;
 37     if(x<=0||y<=0)
 38         return;
 39     if(x==1&&y==1)
 40         return;
 41     if(fx[x][y][k]==0){
 42         out(x-1,y,k);
 43         putchar('D');
 44     }
 45     else{
 46         out(x,y-1,k);
 47         putchar('R');
 48     }
 49 }
 50 int main(){
 51     int n;
 52     scanf("%d",&n);
 53     int num=0;
 54     int numx,numy;
 55     int x;
 56     for(int i=1;i<=n;i++){
 57         for(int j=1;j<=n;j++){
 58             scanf("%d",&x);
 59             if(x==0)
 60                 a[i][j][1]=a[i][j][0]=1,num+=1,numx=i,numy=j;
 61             else{
 62                 a[i][j][1]=a[i][j][0]=0;
 63                 int p=x;
 64                 while(p%2==0){
 65                     a[i][j][0]+=1;
 66                     p/=2;
 67                 }
 68                 p=x;
 69                 while(p%5==0){
 70                     a[i][j][1]+=1;
 71                     p/=5;
 72                 }
 73             }
 74         }
 75     }
 76     int pt;
 77     pt = dfs(n,n,0);
 78     int pt2;
 79     pt2 = dfs(n,n,1);
 80     int k = 0;
 81     if(pt2<pt){
 82         pt=pt2;
 83         k=1;
 84     }
 85     int ans = pt;
 86 /*    for(int i=1;i<=n;i++){
 87         for(int j=1;j<=n;j++){
 88             cout<<dfs(i,j,0)<<" ";
 89         }
 90         cout<<endl;
 91     }
 92     cout<<endl<<endl<<endl;
 93     for(int i=1;i<=n;i++){
 94         for(int j=1;j<=n;j++){
 95             cout<<dfs(i,j,1)<<" ";
 96         }
 97         cout<<endl;
 98     }*/
 99     if(num==0||ans<=1){
100         printf("%d\n",ans);
101         out(n,n,k);
102         return 0;
103     }
104     printf("%d\n",1);
105     for(int i=1;i<numx;i++)
106         putchar('D');
107     for(int i=1;i<numy;i++)
108         putchar('R');
109     for(int i=numx;i<n;i++)
110         putchar('D');
111     for(int i=numy;i<n;i++)
112         putchar('R');
113     return 0;
114 }

 

转载于:https://www.cnblogs.com/xfww/p/8404405.html