POJ3468 A Simple Problem with Integers

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本文介绍了一个涉及整数操作的问题，通过使用线段树数据结构来高效处理区间加法及查询操作。该方法能够快速响应大量查询，适用于解决算法竞赛中的区间更新与求和问题。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 57061		Accepted: 17322
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A₁, A₂, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4

Sample Output

Hint

The sums may exceed the range of 32-bit integers.

线段树区间修改。。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX=100010;
struct seg
{
    int l,r;
    __int64 sum,mark;       //两个附加信息，一个表示和，一个表示增加了多少
}tree[MAX*3];
int num[MAX];
void build(int l,int r,int k)   //建立线段树
{
    tree[k].l=l;
    tree[k].r=r;
    tree[k].mark=0;
    if(l==r)
    {
        tree[k].sum=num[l];
        return;
    }
    int mid=(l+r)>>1;
    build(l,mid,2*k);
    build(mid+1,r,2*k+1);
    tree[k].sum=tree[2*k].sum+tree[2*k+1].sum;
}
void insert(int l,int r,int k,int val)
{
    if(tree[k].l==l&&tree[k].r==r)  //如果节点的区间和要增加的区间完全重合，增加节点的mark
    {
        tree[k].mark+=val;
        return;
    }
    tree[k].sum+=val*(r-l+1);   //不重合，这个值肯定要往下延伸，为了保证不超时，我们只需把这个区间的值加上增加的值，
    //此时这个节点的mark没有变化
    int mid=(tree[k].l+tree[k].r)>>1;
    if(mid>=r)
        insert(l,r,2*k,val);
    else if(mid<l)
        insert(l,r,2*k+1,val);
    else
    {
        insert(l,mid,2*k,val);
        insert(mid+1,r,2*k+1,val);
    }
}
__int64 query(int l,int r,int k)
{
    if(tree[k].l==l&&tree[k].r==r)
        return tree[k].sum+(r-l+1)*tree[k].mark;
    if(tree[k].mark!=0) //如果这个节点的mark不为0，这个节点的子节点肯定还没有mark，所以拓展到后面的子节点
    {
        tree[2*k].mark+=tree[k].mark;
        tree[2*k+1].mark+=tree[k].mark;
        tree[k].sum+=(tree[k].r-tree[k].l+1)*tree[k].mark;
        tree[k].mark=0;
    }
    int mid=(tree[k].l+tree[k].r)>>1;
    if(mid>=r)
        return query(l,r,2*k);
    else if(mid<l)
        return query(l,r,2*k+1);
    else
        return query(l,mid,2*k)+query(mid+1,r,2*k+1);
}
int main()
{
    int n,m,i,x,y,a;
    char s[2];
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)==2)
    {
        for(i=1;i<=n;i++)
            scanf("%d",&num[i]);
        build(1,n,1);
        while(m--)
        {
            scanf("%s",s);
            if(s[0]=='Q')
            {
                scanf("%d%d",&x,&y);
                __int64 ans=query(x,y,1);
                printf("%I64d\n",ans);
            }
            else
            {
                scanf("%d%d%d",&x,&y,&a);
                insert(x,y,1,a);
            }
        }
    }
    return 0;
}