algorithm

import java.

import java.

import java.

Question My thought Nothing… The problem is pretty easy, however it troubled me for a quite long time.

ArrayList // the data type must be a type of class ArrayList<String> a = new ArrayList<String>(); ArrayList<Integer> b = new ArrayList<Integer>(); // the size of the ArrayList arrayList.

The server’s previous technical issues have caused the loss of images and data.

花店橱窗布置 思路 f[i][j]f[i][j]表示前i个花瓶前j个花束的最大美学价值 f[i][j]=max(f[i−1][k],f[i][j])f[i][j]=max(f[i−1][k],f[i][j]) 当然还有另外一种思路(*太强了！！！*)：

define N 1e5 struct bign { int len; int v[N]; // 赋值 bign=bign bign operator = (char* s) { len=strlen(s); memset(v,0,sizeof(v)); for(int i=0;i<len;i++) v[i]=s[len-i-1]-'0'; return *this; } //赋值 bign=int bign operator = (int x) { char s[N]; sprintf(s,"%d",x); return *this=s; } // 高精加 bign operator + (const bign &b) const { bign c; memset(c.

二叉苹果树 思路 $dp[u][j表示节点u留下j条边的最大价值，每一次决策只有三种情况：剪左子树，剪右子树，两个都不剪 剪左边：dp[u][j]=dp[rson][j−1]+v[rson]dp[u][j]=dp[rson][j−1]+v[rson]，同理，剪右边：dp[u][j]=dp[lson][j−1]+v[lson]dp[u][j]=dp[lson][j−1]+v[lson] 两边都不剪：dp[u][j]=dp[lson][j]+dp[rson][k−j−2]dp[u][j]=dp[lson][j]+dp[rson][k−j−2] 代码：记忆化搜索 int f[N][N]; bool t[N][N]; int dp(int u,int k) { if(t[u][k]) return f[u][k]; t[u][k]=1; if(!

The server’s previous technical issues have caused the loss of images and data.