CF 2B The least round way

最新推荐文章于 2024-04-16 22:33:32 发布

XMzhou

最新推荐文章于 2024-04-16 22:33:32 发布

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分类专栏： CF 文章标签： ACM DP

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本文介绍了一种算法，用于在一个n×n的矩阵中找到一条从左上角到右下角的路径，使得该路径上所有数值相乘后的结果中尾随零的数量最少。文章详细解释了算法的实现思路，并提供了完整的代码示例。

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.

Sample test(s)

input
3
1 2 3
4 5 6
7 8 9

output
0
DDRR


题意是给你一个n * n 的矩阵，他的每一步只能向下或向右走，问从左上角到右下角的路径中乘积中零最少的路径，输出路径和零的个数。
思路：由于每个位置只能有其上的位置或其右的位置得到，两者取min即可；
dp[i][j][k]表示从（0 ， 0）到（i ， j）位置的路径中乘积零最小的路径所包含的2数（k = 0）5数（k = 1）
特例是矩阵中有零且有2和5，则必走有零的那条路径。
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>

using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid  , rt << 1
#define rson mid + 1 , r , rt << 1 | 1

#define mk make_pair
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 1000 + 50;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-4
const int MOD = (int)1e9 + 7;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
///#pragma comment(linker, "/STACK:102400000,102400000")
int dp[MAXN][MAXN][2] , a[MAXN][MAXN][2] , n ;
int main()
{
    //ios::sync_with_stdio(false);
#ifdef Online_Judge
    //freopen("in.txt","r",stdin);
//    freopen("out.txt","w",stdout);
#endif // Online_Judge
    int x ,  r;
    bool ok;
    while(~scanf("%d" , &n))
    {
        ok = false;
        FOR(i ,0 , n)FOR(j ,0 , n)
        {
            scanf("%d" , &x);
            if(x == 0)
            {
                ok = true;
                r = i ; x = 10;
            }
            while(x % 2 == 0)
            {
                x /= 2;
                dp[i][j][0] ++;
            }
            while(x % 5 == 0)
            {
                x /= 5;
                dp[i][j][1]++;
            }
            FOR(k , 0 , 2)a[i][j][k] = dp[i][j][k];
        }
        FOR(i , 1 , n)FOR(j , 0 , 2)
        {
            dp[0][i][j] += dp[0][i - 1][j];
            dp[i][0][j] += dp[i - 1][0][j];
        }
        FOR(i , 1 , n)FOR(j , 1 , n)FOR(k , 0 ,2)
        {
            dp[i][j][k] += min(dp[i - 1][j][k] , dp[i][j - 1][k]);
        }
        int k = dp[n - 1][n - 1][1] < dp[n - 1][n - 1][0] ;
        if(ok)
        {
            if(dp[n - 1][n - 1][k] > 1)
            {
                printf("1\n");
                FORR(i , 1 , r)printf("D");
                FOR(i , 0 , n - 1)printf("R");
                FOR(i , r + 1 , n)printf("D");
                continue;
            }
        }
        printf("%d\n" , dp[n - 1][n - 1][k]);
        int i = n - 1 , j = n - 1;
        char s[MAXN * 3];
        int len = 0;
        while(i != 0||j != 0)
        {
            if(i > 0&&dp[i - 1][j][k] + a[i][j][k] == dp[i][j][k])
            {
                s[len++] = 'D';
                i--;
            }
            else
            {
                s[len++] = 'R';
                j--;
            }
        }
        REPP(i ,len - 1 , 0)printf("%c" , s[i]);
        printf("\n");
    }

    return 0;
}