D - A Simple Problem with Integers

最新推荐文章于 2023-11-12 12:05:35 发布

原创最新推荐文章于 2023-11-12 12:05:35 发布 · 431 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#build #numbers #each #64bit #struct #output

数据结构__练习专栏收录该内容

26 篇文章

订阅专栏

本文介绍了一种用于处理区间更新和查询的高效算法。通过构建数据结构，能够在给定的时间限制内，快速地对指定区间内的整数进行加法操作，并询问特定区间的元素之和。算法采用线段树结构，支持复杂操作，同时考虑了32位整数溢出的问题。

Time Limit: 5000MS

Memory Limit: 131072KB

64bit IO Format: %I64d & %I64u

[Submit] [Go Back] [Status]

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.

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include<iostream>
using namespace std;
#define MAXN 102000
struct  node 
{
    int l;
    int r;
    __int64 sum;
    __int64 add;
}t[4*MAXN];
__int64 rr[MAXN];
void build(int id,int l,int r)
{
    int mid = (l+r)/2;
    t[id].l = l;
    t[id].r = r;
    if(r==l)
    {
        t[id].sum = rr[l];
        t[id].add = 0;
    }
    else
    t[id].add = 0;

    if(r>l)
    {
        build(id*2,l,mid);
        build(id*2+1,mid+1,r);
        t[id].sum = t[id*2].sum + t[id*2+1].sum;
    }
}
void change(int id,int start,int end,__int64 num)
{
    if(start==t[id].l && end==t[id].r)
    {
        t[id].add += num;
        return ;
    }
    t[id].sum+= (end-start+1)*num;
    int mid = (t[id].l + t[id].r)/2;
    if(t[id].add!=0)
    {
        t[id].sum += (t[id].r-t[id].l+1)*t[id].add;
        change(id*2,t[id].l,mid,t[id].add);
        change(id*2+1,mid+1,t[id].r,t[id].add);
        t[id].add = 0;
    }
    if(mid>=end)
    {
         change(id*2,start,end,num);
         
    }
    else if(mid+1<=start)
    {
          change(id*2+1,start,end,num);
    }
    else
    {
        change(id*2,start,mid,num);
        change(id*2+1,mid+1,end,num);
    }
}

__int64 ans;
void find(int id,int start,int end)
{
    if (t[id].l==start && t[id].r == end)
    {
        ans+=t[id].sum+(t[id].r-t[id].l+1)*t[id].add;
        return;
    }
    if (t[id].add!=0)
    {
        t[id].sum+= (t[id].r-t[id].l+1)*t[id].add;
        t[id*2].add+=t[id].add;
        t[id*2+1].add+=t[id].add;
        t[id].add = 0;
    }

    int mid = (t[id].l+t[id].r)/2;
    if(mid>=end)
    find(id*2,start,end);
    else if(start>=mid+1)
    find(id*2+1,start,end);
    else
    {
        find(id*2,start,mid);
        find(id*2+1,mid+1,end);
    }

}

 

int main()
{
	int n,m;
    int i,j;
    char str[10];
    int start,end;
    __int64 num;
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        for (i=1;i<=n;i++)
        {
            scanf("%I64d",&rr[i]);
        }
        build(1,1,n);
        for (i=0;i<m;i++)
        {
            scanf("%s",str);
            if (strcmp(str,"C")==0)
            {
                scanf("%d%d%I64d",&start,&end,&num);
                change(1,start,end,num);
            }
            else
            {
                scanf("%d%d",&start,&end);
                ans = 0;
                find(1,start,end);
                printf("%I64d\n",ans);
            }
        }
    }
    return 0;
}