20190918 Meituan, Xiaoma Zhixing

Meituan

1.

 

n=int(input().strip())

res = 1
cnt = 0
a = list(range(n+1))
for i in range(5,n+1,5):
    while a[i]%5==0:
        a[i]//=5
        cnt+=1

i = 2
while i//2<=cnt:
    a[i]//=2
    i+=2

for i in range(2,n+1):
    res*=a[i]
    while res%10==0:
        res//=10
    res%=10
print(res)

2.

 

n=int(input().strip())

t = 1
for i in range(1,n+1):
    t*=i
    t%=(10**6+3)
print(t)

3.

This law example cannot pass, but it can pass 73%, dfs

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;

void dfs(vector<vector<int>>& graph,vector<bool>& vis, int n,int pos,int end) {
	if (vis[pos] == true) return;
	if (pos == end) {
		vis[end] = true;
		return;
	}
	vis[pos] = true;
	for (int i = 1; i <= n; ++i) {
		if (graph[pos][i] == 1 && vis[i] == false) dfs(graph, vis, n, i, end);
	}
}

int main() {
	int n, m;
	cin >> n >> m;
	vector<vector<int>> graph(n + 1, vector<int>(n + 1, 0));
	int u, v;
	for (int i = 0; i < m; ++i) {
		cin >> u >> v;
		graph[u][v] = 1;
		graph[v][u] = 1;
	}
	int s, t;
	cin >> s >> t;
	vector<bool> vis(n + 1, false);
	dfs(graph, vis, n, s, t);
	for (int i = 1; i <= n; ++i) {
		if (!vis[i]) cout << i << " ";
	}
	return 0;
}

Xiaoma Zhixing

1.

n numbers, find the truncated average for each consecutive k numbers, and find the largest truncated average

90%

n,k= map(int, input().strip().split())
arr=list(map(int, input().strip().split()))
ans=0
sum1=sum(arr[0:k])
max1=max(arr[0:k])
min1=min(arr[0:k])
mean=(sum1-min1-max1)/(k-2)
ans=mean
for i in range(n-k):
    sum1=sum1-arr[i]+arr[i+k]
    if arr[i]==max1:
        max1=max(arr[i+1:i+k+1])
    if arr[i]==min1:
        min1=min(arr[i+1:i+k+1])
    mean=(sum1-min1-max1)/(k-2)
    if mean>ans:
        ans=mean
    #print(sum1,max1,min1,mean)
print(ans)

2.

There are n stones in a row from left to right, and the color of the i-th stone is ci. Now Little Ho is on the first rock, and he wants to jump to the nth rock. There are two jumping methods for each step:

1. Jump from the i-th stone to the i+1-th stone

2. Jump to the first stone on the right with the same color as the current stone (if it exists)

Now he wants to know that it takes at least a few steps to jump from the first rock to the nth rock

Enter
a positive integer n in the first line

N positive integers in the second line, representing ci

1 ≤ n, ci ≤ 2 × 105

Output
output requires at least a few steps to jump

Sample input
6
1 2 1 3 4 2
Sample output
2

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n;
int c[200010];//输入数组 
int vis[200010];//记录数组中每个颜色出现的次数 
int dp[200010];//统计每到一步需要走的最少步数 
int a[200010];//我的循环从左到右，标记走到当前位置时，前面与之是相同颜色的最右边的位置i 
int main()
{
	cin>>n;
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=n;i++)
	{
		cin>>c[i];
		vis[c[i]]++;//统计颜色出现的次数 
	}
	memset(dp,0,sizeof(dp));
	memset(a,0,sizeof(a));
	dp[0]=-1;//很关键的初始化，因为下面的dp中第一个位置到后面位置可能需要用到dp[0] 
	a[c[1]]=1;//第一个元素的位置在1号位 
	for(int i=2;i<=n;i++)
	{
		int k=a[c[i]];//当前元素的位置 
		if(vis[c[i]]>=2 && dp[k]>=0) dp[i]=min(dp[i-1]+1,dp[k]+1);//满足该种颜色出现两次（含）以上并且标记k不是第一个位置 
		else dp[i]=dp[i-1]+1;
		a[c[i]]=i;//更新当前标记 
	}
	cout<<dp[n]<<endl;
	return 0;
}

There are two more questions, the super complicated one