codeforces703D 【Mishka and Interesting sum】

本文深入探讨了求解区间内出现偶数次数值的异或和问题,提出了一种结合线段树与前缀和的高效算法。通过维护区间内不同数值的异或和,实现了快速查询和更新,适用于大数据范围的问题。

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首先,题意就不多说了;

观察题目,我们发先要求的是出现偶数个数的数的亦或和,根据亦或的性质我们知道,一个数如果被亦或偶数次,那么就是它本身,
说以稍加思索~~(很容易)~~可以发现,
区间出现偶数次的数的异或和=区间异或和^区间所有不同数的异或和。所以我们只需用线段树维护区间内所有不同数的亦或和,而区间亦或和,具体可参见sp3267,只需在前者上做一些改动,,而区间亦或和,这可以用前缀和来完成。

ps:由于此题数据范围较大~~(我懒得写离散化)~~,于是用map来记录每个数上一次出现的位置

废话不多说,上代码:

#include <bits/stdc++.h>
#define ll long long

using namespace std;

const int N=4e6+5;//线段树的4倍空间 
map <int,int> pos;//用于记录每个数上一次出现的位置 
int L[N],R[N];
int letter[N]; 
int dp[N];//区间亦或和 
int sum[N];//线段树 
int ans[N];//记录答案- 
int n,m;

struct node
{
    int l,r;//区间 
    int cnt;//读入时的位置 
    bool operator <(const node &b)const
    {
        return r<b.r;
    }
}a[N];
//线段树裸模板 
void build(int l,int r,int now)
{
    L[now]=l;
    R[now]=r;
    sum[now]=0;
    if (l==r)return;
    int mid=(l+r)>>1;
    build(l,mid,now<<1);
    build(mid+1,r,now<<1|1);
    sum[now]=sum[now<<1]^sum[now<<1|1];
}

void update(int l,int r,int now,int d)
{
    if (L[now]==l&&R[now]==r)
    {
        sum[now]^=d;
        return;
    }
    int mid=(L[now]+R[now])>>1;
    if (r<=mid)update(l,r,now<<1,d);
    else if (l>mid)update(l,r,now<<1|1,d);
    sum[now]=sum[now<<1]^sum[now<<1|1];
}

int query(int l,int r,int now)
{
    if (L[now]==l&&r==R[now])return sum[now];
    int mid=(L[now]+R[now])>>1;
    if (r<=mid)return query(l,r,now<<1);
    else if (l>mid)return query(l,r,now<<1|1);
    else return query(l,mid,now<<1)^query(mid+1,r,now<<1|1);
    sum[now]=sum[now<<1]^sum[now<<1|1];
}

int main()
{
    scanf("%d",&n);
    build(1,n,1);
    for (int i=1;i<=n;i++)
    {
        scanf("%d",&letter[i]);
        dp[i]=dp[i-1]^letter[i];//区间亦或和 
    }	
    scanf("%d",&m);
    for (int i=1;i<=m;i++)
    {
        scanf("%d%d",&a[i].l,&a[i].r);
        a[i].cnt=i;//小trick 
    }
    sort(a+1,a+1+m);
    int last=0;
    for (int i=1;i<=m;i++)
    {
        while(last<a[i].r)
        {
            last++;
            if (pos[letter[last]])
                update(pos[letter[last]],pos[letter[last]],1,letter[last]);//其实是本质是单点加 
            update(last,last,1,letter[last]);
            pos[letter[last]]=last;
        }
        ans[a[i].cnt]=query(a[i].l,a[i].r,1)^dp[a[i].r]^dp[a[i].l-1]; 
    }
    for (int i=1;i<=m;i++)printf("%d\n",ans[i]);
    return 0;
}
### Codeforces 1487D Problem Solution The problem described involves determining the maximum amount of a product that can be created from given quantities of ingredients under an idealized production process. For this specific case on Codeforces with problem number 1487D, while direct details about this exact question are not provided here, similar problems often involve resource allocation or limiting reagent type calculations. For instance, when faced with such constraints-based questions where multiple resources contribute to producing one unit of output but at different ratios, finding the bottleneck becomes crucial. In another context related to crafting items using various materials, it was determined that the formula `min(a[0],a[1],a[2]/2,a[3]/7,a[4]/4)` could represent how these limits interact[^1]. However, applying this directly without knowing specifics like what each array element represents in relation to the actual requirements for creating "philosophical stones" as mentioned would require adjustments based upon the precise conditions outlined within 1487D itself. To solve or discuss solutions effectively regarding Codeforces' challenge numbered 1487D: - Carefully read through all aspects presented by the contest organizers. - Identify which ingredient or component acts as the primary constraint towards achieving full capacity utilization. - Implement logic reflecting those relationships accurately; typically involving loops, conditionals, and possibly dynamic programming depending on complexity level required beyond simple minimum value determination across adjusted inputs. ```cpp #include <iostream> #include <vector> using namespace std; int main() { int n; cin >> n; vector<long long> a(n); for(int i=0;i<n;++i){ cin>>a[i]; } // Assuming indices correspond appropriately per problem statement's ratio requirement cout << min({a[0], a[1], a[2]/2LL, a[3]/7LL, a[4]/4LL}) << endl; } ``` --related questions-- 1. How does identifying bottlenecks help optimize algorithms solving constrained optimization problems? 2. What strategies should contestants adopt when translating mathematical formulas into code during competitive coding events? 3. Can you explain why understanding input-output relations is critical before implementing any algorithmic approach? 4. In what ways do prefix-suffix-middle frameworks enhance model training efficiency outside of just tokenization improvements? 5. Why might adjusting sample proportions specifically benefit models designed for tasks requiring both strong linguistic comprehension alongside logical reasoning skills?
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