在网上看了一道笔试题，就是简单的dp，准备一下字节跳动夏令营的笔试
输入16进制数(1,2…f…)组成的矩阵，由左上角开始，到右下角，只能向右或下走，找出另走过的数的乘积的16进制数后缀0最少的方法。

#include<iostream>

using namespace std;

int nn[100][100];
int dp[100][100];

void myin(int n, int m)
{
	char c;
	for(int i = 0; i<m; i++)
	{
		for(int j = 0; j<n; j++)
		{
			cin >> hex >> nn[i][j];
		}
	}
}

int pls(int i, int j, int a, int b)
{
	int plss = 0;
	int miu = nn[i-a][j-b] * nn[i][j];
	while(1)
	{
		if(miu % 16 == 0)
		{
			plss++;
			miu /= 16;
		}
		else
		 break;
	}
	nn[i][j] = miu;
	return plss;
}

int main()
{
	int n,m;
	cin >> n >> m;
	myin(n, m);
	for(int i = 0; i<m; i++)
	{
		for(int j = 0; j<n; j++)
		{
			if(i == 0 && j == 0)
				dp[i][j] = 0;
			else if(i == 0)
				dp[i][j] = dp[i][j - 1] + pls(i,j,0,1);
			else if(j == 0)
				dp[i][j] = dp[i - 1][j] + pls(i,j,1,0);
			else
				dp[i][j] = dp[i][j - 1]<dp[i - 1][j]?(dp[i][j - 1] + pls(i,j,0,1)):(dp[i - 1][j] + pls(i,j,1,0));
		}
	}
	printf("乘积矩阵：\n");
	for(int i = 0; i<m; i++)
	{
		for(int j = 0; j<n; j++)
		{
			printf("%x ",nn[i][j]);
		}
		printf("\n");
	}
	printf("状态矩阵：\n");
	for(int i = 0; i<m; i++)
	{
		for(int j = 0; j<n; j++)
		{
			printf("%d ",dp[i][j]);
		}
		printf("\n");
	}
	printf("最小末尾0：%d\n",dp[n-1][m-1]);
}