Codeforces 813E Army Creation [主席树]

本文介绍了一种使用主席树解决区间内元素计数问题的方法。通过构建主席树并结合差分思想,可以有效地处理区间内相同数字最多选择k个的情况。文章提供了详细的AC代码实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

题意:给你长度为n的序列,有q个询问,询问[l,r]区间内相同数字最多选k个的总个数。

题解:建立关于位置的主席树,利用差分的思想,对于新加入的一个数,当前位置+1,并在之前第i-k个位置-1,最后询问root[r]中区间[l,r]的总和。

AC代码:

#include<stdio.h>  
#include<vector>  
#include<algorithm>  
#define N 100005  
using namespace std;  
vector<int>vt[N];  
int tree[N*36],lchild[N*36],rchild[N*36],root[N],tot;  
void insert(int last,int cur,int x,int L,int R,int k)  
{  
    tree[cur]=tree[last]+k;  
    lchild[cur]=lchild[last];  
    rchild[cur]=rchild[last];  
    if(L==R)return ;  
    int mid=L+R>>1;  
    if(x<=mid)insert(lchild[last],lchild[cur]=++tot,x,L,mid,k);  
    else insert(rchild[last],rchild[cur]=++tot,x,mid+1,R,k);  
}  
int query(int l,int r,int L,int R,int root)  
{  
    if(l<=L&&R<=r)return tree[root];  
    int mid=L+R>>1;  
    if(r<=mid)return query(l,r,L,mid,lchild[root]);  
    else if(l>mid)return query(l,r,mid+1,R,rchild[root]);  
    else return query(l,mid,L,mid,lchild[root])+query(mid+1,r,mid+1,R,rchild[root]);  
}  
int main()  
{  
    int n,k,q,x;  
    scanf("%d%d",&n,&k);  
    for(int i=1;i<=n;i++)  
    {  
        scanf("%d",&x);  
        vt[x].push_back(i);  
        insert(root[i-1],root[i]=++tot,i,1,n,1);  
        if(vt[x].size()>k)
		{
			int now=++tot;
			insert(root[i],now,vt[x][vt[x].size()-k-1],1,n,-1);
			root[i]=now;  
		}
    }  
    int last=0;  
    scanf("%d",&q);  
    while(q--)  
    {  
        int l,r;  
        scanf("%d%d",&l,&r);  
        l=(l+last)%n+1;  
        r=(r+last)%n+1;  
        if(l>r)swap(l,r);  
        last=query(l,r,1,n,root[r]);  
        printf("%d\n",last);  
    }  
}  




### Codeforces 887E Problem Solution and Discussion The problem **887E - The Great Game** on Codeforces involves a strategic game between two players who take turns to perform operations under specific rules. To tackle this challenge effectively, understanding both dynamic programming (DP) techniques and bitwise manipulation is crucial. #### Dynamic Programming Approach One effective method to approach this problem utilizes DP with memoization. By defining `dp[i][j]` as the optimal result when starting from state `(i,j)` where `i` represents current position and `j` indicates some status flag related to previous moves: ```cpp #include <bits/stdc++.h> using namespace std; const int MAXN = ...; // Define based on constraints int dp[MAXN][2]; // Function to calculate minimum steps using top-down DP int minSteps(int pos, bool prevMoveType) { if (pos >= N) return 0; if (dp[pos][prevMoveType] != -1) return dp[pos][prevMoveType]; int res = INT_MAX; // Try all possible next positions and update 'res' for (...) { /* Logic here */ } dp[pos][prevMoveType] = res; return res; } ``` This code snippet outlines how one might structure a solution involving recursive calls combined with caching results through an array named `dp`. #### Bitwise Operations Insight Another critical aspect lies within efficiently handling large integers via bitwise operators instead of arithmetic ones whenever applicable. This optimization can significantly reduce computation time especially given tight limits often found in competitive coding challenges like those hosted by platforms such as Codeforces[^1]. For detailed discussions about similar problems or more insights into solving strategies specifically tailored towards contest preparation, visiting forums dedicated to algorithmic contests would be beneficial. Websites associated directly with Codeforces offer rich resources including editorials written after each round which provide comprehensive explanations alongside alternative approaches taken by successful contestants during live events. --related questions-- 1. What are common pitfalls encountered while implementing dynamic programming solutions? 2. How does bit manipulation improve performance in algorithms dealing with integer values? 3. Can you recommend any online communities focused on discussing competitive programming tactics? 4. Are there particular patterns that frequently appear across different levels of difficulty within Codeforces contests?
评论 1
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值