线段树（树状数组）hdu4267

最新推荐文章于 2021-06-09 19:17:31 发布

原创最新推荐文章于 2021-06-09 19:17:31 发布 · 517 阅读

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本文介绍了一个在线编程平台的使用方法，包括在线练习、教学、竞赛等内容。重点讲解了如何解决一道涉及区间操作和元素修改的算法题，通过分组和线段树/树状数组等数据结构实现高效处理。

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A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3544 Accepted Submission(s): 1094

Problem Description

Let A1, A2, ... , AN be N elements. You need to deal with two kinds of operations. One type of operation is to add a given number to a few numbers in a given interval. The other is to query the value of some element.

Input

There are a lot of test cases.
The first line contains an integer N. (1 <= N <= 50000)
The second line contains N numbers which are the initial values of A1, A2, ... , AN. (-10,000,000 <= the initial value of Ai <= 10,000,000)
The third line contains an integer Q. (1 <= Q <= 50000)
Each of the following Q lines represents an operation.
"1 a b k c" means adding c to each of Ai which satisfies a <= i <= b and (i - a) % k == 0. (1 <= a <= b <= N, 1 <= k <= 10, -1,000 <= c <= 1,000)
"2 a" means querying the value of Aa. (1 <= a <= N)

Output

For each test case, output several lines to answer all query operations.

Sample Input

Sample Output

又是一道神奇的题目。。。

这个题是对区间内不同的数加上一个值，但线段树只能维护连续的区间，所以这里我们按照k进行分组，根据(i-a)%k==0 --> i%k=a%k。

分组后一共有55中情况：

1,2,3,4,5......
1,3,5,7,9......
2,4,6,8,10....
1,4,7,10,13...
2,5,9,12,15...
3,6,10,13,16...

这样输入a，b，就可以定位到tree[k][a%k]这一组，然后对组内进行成段更新，线段树和树状数组都可以。

Tips：

树状数组的优势是方便动态求值和更新..

可惜树状数组是单点更新

倒是有个方法可以快速成段更新

就是在区间[a, b]内更新+x就在a的位置+x 然后在b+1的位置-x

求某点a的值就是求数组中1~a的和..

树状数组代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=50010;
int n,a[maxn];

struct BIT
{
    int val[maxn];
    void clear(){memset(val,0,sizeof(val));}
    int lowbit(int x)
    {
        return x&(-x);
    }
    void update(int s,int e,int x)
    {
        while(e>0)
        {
            val[e]+=x;
            e-=lowbit(e);
        }
        s--;
        while(s>0)
        {
            val[s]-=x;
            s-=lowbit(s);
        }
    }
    int getsum(int x)
    {
        int sum=0;
        while(x<maxn)
        {
            sum+=val[x];
            x+=lowbit(x);
        }
        return sum;
    }
}tree[12][12];

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        for(int i=0;i<=10;i++)
            for(int j=0;j<=10;j++)tree[i][j].clear();
        int q;
        scanf("%d",&q);
        while(q--)
        {
            int op,l,r,k,x;
            scanf("%d",&op);
            if(op==1)
            {
                scanf("%d%d%d%d",&l,&r,&k,&x);
                tree[k][l%k].update(l/k+1,l/k+(r-l)/k+1,x);
            }
            else
            {
                scanf("%d",&x);
                int sum=0;
                for(int i=1;i<=10;i++)
                    sum+=tree[i][x%i].getsum(x/i+1);
                printf("%d\n",sum+a[x]);
            }
        }
    }
    return 0;
}

线段树代码（别人的）：

#include<algorithm>
#include<cstdio>
#include<string.h>
#include<vector>
using namespace std;
const int N=50010;
int sum[N<<2][55];
int col[N<<2],res[N];
#define lson l,mid, rt << 1
#define rson  mid+1,r,rt << 1 | 1
int mod,LL;
int shunxu[11][11];
void PushDown(int rt)
{
    if(col[rt])
    {
        col[rt<<1] += col[rt];
        col[rt<<1|1] += col[rt];
        col[rt]=0;
        for(int i=0; i<55; i++)
        {
            sum[rt<<1][i] += sum[rt][i];
            sum[rt<<1|1][i] += sum[rt][i] ;
            sum[rt][i]=0;
        }

    }
}
void build(int l,int r,int rt)
{
    col[rt]=0;
    memset(sum[rt],0,sizeof(sum[rt]));
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
}
void update(int L,int R,int val,int l,int r,int rt)
{
    if(l==L&&r==R)
    {
        col[rt]+=val;
        sum[rt][shunxu[mod][LL%mod]]+=val;
        return ;
    }
    PushDown(rt);
    int mid=(l+r)>>1;
    if(R<=mid)update(L,R,val,lson);
    else if(L>mid)update(L,R,val,rson);
    else
    {
        update(L,mid,val,lson);
        update(mid+1,R,val,rson);
    }
}
int query(int l,int r,int rt,int m)
{

    if(l==r)
    {
        int d=0;
        for(int i=1; i<=10; i++)
            d += sum[rt][shunxu[i][m%i]];
        return d+res[l];
    }
    int mid=(l+r)>>1;
    PushDown(rt);
    if(m<=mid)
        return query(lson,m);
    else return query(rson,m);
}
int main()
{
    int n,q,cc=0;
    for(int i=1; i<=10; i++)
        for(int j=0; j<i; j++)
            shunxu[i][j]=cc++;
    while(~scanf("%d",&n))
    {
        build(1,n,1);
        int q;
        for(int i=1; i<=n; i++)
            scanf("%d",&res[i]);
        scanf("%d",&q);

        int kk,aa,RR,k,val;
        while(q--)
        {
            scanf("%d",&kk);
            if(kk==2)
            {
                scanf("%d",&aa);
                printf("%d\n",query(1,n,1,aa));
            }
            else if(kk==1)
            {
                scanf("%d%d%d%d",&LL,&RR,&mod,&val);
                update(LL,RR,val,1,n,1);
            }
        }
    }
    return 0;
}