B. The least round way

最新推荐文章于 2024-09-25 00:00:06 发布

UFO_good

最新推荐文章于 2024-09-25 00:00:06 发布

阅读量422

点赞数

CC 4.0 BY-SA版权

分类专栏： [Codeforces]

本文链接：https://blog.youkuaiyun.com/ljlsn/article/details/63691678

[Codeforces] 专栏收录该内容

5 篇文章

订阅专栏

本文介绍了一种算法，用于在一个非负整数构成的矩阵中找到一条从左上角到右下角的路径，使得路径上的数值乘积结束时包含的零最少。通过动态规划方法分别计算2和5因子的数量，并考虑特殊情况处理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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
思路：利用dp思想，寻找2最少与5最少，最后也需要考虑特殊情况当数字为0时，输入时记得用scanf输入。

#include <iostream>
#include <cstring>
#include <cmath>
#include <cstdio>
using namespace std;

const int MAX = 1010;
int n;

struct data{
    int value;
    int num_2;
    int num_5;
    char last_path[2];  //0 : "2_path" 1: "5_path"
}data[MAX][MAX];

int buffer_dp[MAX][MAX]={
    0
};


int count(int number,int base){
    int count = 0;
    if(number==0)
        return count;      
    while(number%base==0){
        count++;
        number=number/base;
    }
    return count;
}

void path_2(){
    buffer_dp[0][0] = data[0][0].num_2;
    for(int j = 1;j<n;++j){
        buffer_dp[0][j] = buffer_dp[0][j-1] + data[0][j].num_2;
        data[0][j].last_path[0] = 'R';
    }
    for(int i = 1;i<n;++i){
        buffer_dp[i][0] = buffer_dp[i-1][0] + data[i][0].num_2;
        data[i][0].last_path[0] = 'D';
    }

    for(int i=1;i<n;++i){
        for(int j = 1;j<n;++j){
            if(buffer_dp[i-1][j]<buffer_dp[i][j-1]){
                buffer_dp[i][j] = buffer_dp[i-1][j] + data[i][j].num_2;
                data[i][j].last_path[0] = 'D';
            }   
            else{
                buffer_dp[i][j] = buffer_dp[i][j-1] + data[i][j].num_2;
                data[i][j].last_path[0] = 'R';
            }
        }
    }             
}

void path_5(){
    buffer_dp[0][0] = data[0][0].num_5;
    for(int j = 1;j<n;++j){
        buffer_dp[0][j] = buffer_dp[0][j-1] + data[0][j].num_5;
        data[0][j].last_path[1] = 'R';
    }
    for(int i = 1;i<n;++i){
        buffer_dp[i][0] = buffer_dp[i-1][0] + data[i][0].num_5;
        data[i][0].last_path[1] = 'D';
    }

    for(int i=1;i<n;++i){
        for(int j = 1;j<n;++j){
            if(buffer_dp[i-1][j]<buffer_dp[i][j-1]){
                buffer_dp[i][j] = buffer_dp[i-1][j] + data[i][j].num_5;
                data[i][j].last_path[1] = 'D';
            }   
            else{
                buffer_dp[i][j] = buffer_dp[i][j-1] + data[i][j].num_5;
                data[i][j].last_path[1] = 'R';
            }
        }
    }             
}

void Print(int i,int j,int index){

    if( i==0 && j==0 ){
        return ;
    }

    char direction = data[i][j].last_path[index];


    if( direction == 'D')
        Print(i-1,j,index);
    else
        Print(i,j-1,index);
    cout<<direction;
}

void Print_zero(int zero_x,int zero_y){

    for(int i = 0;i<zero_x;++i)
        cout<<"D";
    for(int j = 0;j<zero_y;++j)
        cout<<"R";
    for(int i = zero_x+1 ;i<n;++i)
        cout<<"D";
    for(int j = zero_y+1 ;j<n;++j)
        cout<<"R";

}


int main(){

    bool zero = false;
    int zero_x ,zero_y;
    int value;
    cin>>n;
    for(int i=0;i<n;++i){
        for(int j=0;j<n;++j){
            scanf("%d",&value);
            data[i][j].value = value;

            if( value == 0){
                zero = true;
                zero_x = i;
                zero_y = j;
            }

            data[i][j].value = value;
            data[i][j].num_2 = count(value,2);
            data[i][j].num_5 = count(value,5);
        }
    }
    path_2();
    int min_2 = buffer_dp[n-1][n-1];
    if(min_2==0){
        cout<<min_2<<endl,Print(n-1,n-1,0);
        cout<<endl;
        return 0;
    }



    path_5();
    int min_5 = buffer_dp[n-1][n-1];
    if(min_5==0){
        cout<<min_5<<endl,Print(n-1,n-1,1);
        cout<<endl;
        return 0;
    }


    if(!zero)
        min_2 < min_5?(cout<<min_2<<endl,Print(n-1,n-1,0)):(cout<<min_5<<endl,Print(n-1,n-1,1));
    else{
        if(1<=min(min_2,min_5))
            cout<<1<<endl,Print_zero(zero_x,zero_y);
        else
            min_2 < min_5?(cout<<min_2<<endl,Print(n-1,n-1,0)):(cout<<min_5<<endl,Print(n-1,n-1,1));
    }   
    cout<<endl;

    return 0;
}