Description
有一个n*n的正整数矩阵，要你求一条从第一行第一列的格子到第n行第n列的路，使得你走过的格子里面的数乘起来的值末尾的零的个数最小。输出最小个数。
Input
第一行包括1个数n。
接下来n行每行n个数字。
Output
一个数字表示末尾零最小个数。
Sample Input
3
1 2 3
4 5 6
7 8 9
Sample Output
0
因为都是正数，对这个来说仅仅需统计最少的2或5就可以。相对简单。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF=0x3f3f3f3f;
const int maxn=1010;
int dp[maxn][maxn][2];
int path[maxn][maxn][2];
int mp[maxn][maxn][2];
int n;
void print(int i,int j)
{
if(i==1&&j==1)
return ;
print(path[i][j][0],path[i][j][1]);
if(i-path[i][j][0]==1&&j==path[i][j][1]) printf("%c",'D');
else printf("%c",'R');
}
int main()
{
int x,cnt1,cnt2,temp;
while(~scanf("%d",&n))
{
REPF(i,1,n)
{
REPF(j,1,n)
{
scanf("%d",&x);
cnt1=cnt2=0;temp=x;
while(temp%2==0)
{
temp/=2;
cnt1++;
}
while(x%5==0)
{
x/=5;
cnt2++;
}
mp[i][j][0]=cnt1;
mp[i][j][1]=cnt2;
}
}
CLEAR(dp,INF);
dp[1][1][0]=mp[1][1][0];
dp[1][1][1]=mp[1][1][1];
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
for(int k=0;k<2;k++)//0:2 1:5
{
if(i>1&&dp[i][j][k]>dp[i-1][j][k]+mp[i][j][k])
{
dp[i][j][k]=dp[i-1][j][k]+mp[i][j][k];
path[i][j][0]=i-1;path[i][j][1]=j;
}
if(j>1&&dp[i][j][k]>dp[i][j-1][k]+mp[i][j][k])
{
dp[i][j][k]=dp[i][j-1][k]+mp[i][j][k];
path[i][j][0]=i;path[i][j][1]=j-1;
}
}
}
}
printf("%d
",min(dp[n][n][0],dp[n][n][1]));
// print(n,n);
// puts("");
}
return 0;
}
/*再看Codeforces 2B:
Description
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 exceeding109).
Output
In the first line print the least number of trailing zeros. In the second line print the correspondent way itself.
Sample Input
Input
3 1 2 3 4 5 6 7 8 9
Output
0 DDRR
Source
不仅要输出路径。并且矩阵中还带了0，这就是麻烦的地方。题解：对于要输出的路径。记录前面的一个状态就可以。对于0的处理，假设到终点的2或
5的个数大于等于1了，而矩阵中含0。这时候就是直接答案就是1个0。路径仅仅需找到随意
一个0所在的行列输出就可以。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF=0x3f3f3f3f;
const int maxn=1010;
int dp[maxn][maxn][2];
int path1[maxn][maxn][2];
int path2[maxn][maxn][2];
int mp[maxn][maxn][2];
int n;
void print1(int i,int j)
{
if(i==1&&j==1)
return ;
print1(path1[i][j][0],path1[i][j][1]);
if(i-path1[i][j][0]==1&&j==path1[i][j][1]) printf("%c",'D');
else printf("%c",'R');
}
void print2(int i,int j)
{
if(i==1&&j==1)
return ;
print2(path2[i][j][0],path2[i][j][1]);
if(i-path2[i][j][0]==1&&j==path2[i][j][1]) printf("%c",'D');
else printf("%c",'R');
}
int main()
{
int x,cnt1,cnt2,temp,ans;
int sx,sy;
while(~scanf("%d",&n))
{
int flag=1;
CLEAR(mp,0);
REPF(i,1,n)
{
REPF(j,1,n)
{
scanf("%d",&x);
cnt1=cnt2=0;temp=x;
if(x==0)
{
mp[i][j][0]=mp[i][j][1]=1;
sx=i;sy=j;flag=0;continue;
}
while(temp%2==0)
{
temp/=2;
cnt1++;
}
while(x%5==0)
{
x/=5;
cnt2++;
}
mp[i][j][0]=cnt1;
mp[i][j][1]=cnt2;
}
}
CLEAR(dp,INF);
dp[1][1][0]=mp[1][1][0];
dp[1][1][1]=mp[1][1][1];
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
for(int k=0;k<2;k++)//0:2 1:5
{
if(i>1&&dp[i][j][k]>dp[i-1][j][k]+mp[i][j][k])
{
dp[i][j][k]=dp[i-1][j][k]+mp[i][j][k];
if(!k)
{
path1[i][j][0]=i-1;
path1[i][j][1]=j;
}
else
{
path2[i][j][0]=i-1;
path2[i][j][1]=j;
}
}
if(j>1&&dp[i][j][k]>dp[i][j-1][k]+mp[i][j][k])
{
dp[i][j][k]=dp[i][j-1][k]+mp[i][j][k];
if(!k)
{
path1[i][j][0]=i;
path1[i][j][1]=j-1;
}
else
{
path2[i][j][0]=i;
path2[i][j][1]=j-1;
}
}
}
}
}
ans=min(dp[n][n][0],dp[n][n][1]);
if(ans>=1&&!flag)
{
printf("%d
",1);
for(int i=0;i<sx-1;i++)
printf("%c",'D');
for(int i=0;i<sy-1;i++)
printf("%c",'R');
for(int i=sx;i<n;i++)
printf("%c",'D');
for(int i=sy;i<n;i++)
printf("%c",'R');
puts("");
continue;
}
printf("%d
",ans);
if(ans==dp[n][n][0]) print1(n,n);
else print2(n,n);
puts("");
}
return 0;
}
/*
3
2 2 2
2 2 2
5 5 5
*/