A Simple Problem with Integers（线段树做的我想哭）-优快云博客

本文链接：https://blog.youkuaiyun.com/SDUTyangkun/article/details/53587192

本文介绍了一个涉及区间加法操作及区间求和查询的问题，并提供了一段使用线段树实现的C++代码示例。该算法能够高效处理大量整数的区间修改与查询任务。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 101173		Accepted: 31557
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

真的  做的我想哭

#include<stdio.h>
#include<string.h>
using namespace std;
const int maxn = 100006;
int a[maxn];
long long  ans;
struct node
{
    long long data;
    int l,r;
    long long  add;
}tree[maxn<<2];
void build( int  v, int l, int r)
{
    tree[v].l = l;
    tree[v].r = r;
    tree[v].add = 0;
    if( l == r)
    {

        tree[v].data = a[r]; return;
    }
    int mid = ( l + r) / 2;
    build( v*2, l, mid);
    build( v*2+1, mid+1, r);
    tree[v].data = tree[2*v].data + tree[2*v+1].data;
}

void update( int v, int l, int r, int m)
{
    if(tree[v].l == l && tree[v].r == r)
    {
        tree[v].data += ( r - l + 1)*m;
        tree[v].add += m;
        return;
    }
    if(tree[v].add)
    {
        tree[2*v].data+= ( tree[2*v].r - tree[2*v].l + 1)*tree[v].add;
        tree[2*v].add += tree[v].add;
        tree[2*v+1].data+= ( tree[2*v+1].r - tree[2*v+1].l + 1)*tree[v].add;
        tree[2*v+1].add += tree[v].add;
        tree[v].add = 0;
    }
    int mid = ( tree[v].l + tree[v].r ) / 2;

    if( r <= mid) update( v * 2,l, r, m);

    else
    {
          if( l > mid) update( v * 2 + 1, l, r, m);
          else
          {
               update( v*2, l, mid, m);
               update( v*2+1, mid+1, r, m);
          }

    }
}
void query( int v, int l, int r)
{
    if(tree[v].l == l && tree[v].r == r)
    {

        ans += tree[v].data;
        return;
    }

    if(tree[v].add)
    {
        tree[2*v].data+= ( tree[2*v].r - tree[2*v].l + 1)*tree[v].add;
        tree[2*v].add += tree[v].add;
        tree[2*v+1].data+= ( tree[2*v+1].r - tree[2*v+1].l + 1)*tree[v].add;
        tree[2*v+1].add += tree[v].add;
        tree[v].add = 0;
    }
    int mid = ( tree[v].l + tree[v].r)/2;
    if( r <= mid ) query(v*2, l, r);
    else
    {
         if( l > mid ) query( v*2+1, l, r);
        else
         {
            query(2*v, l ,mid);
            query(2*v+1, mid+1, r);
        }
    }

}


int main()
{
    int n,m;
    int i,j;
       while(~scanf("%d%d", &n, &m))
   {
       for( i = 1; i <= n; i++)
       {
           scanf("%d" ,&a[i]);
       }
       build(1, 1, n);
       int l,r,w;
       char s[10];
       while(m--)
       {
           scanf("%s", s);
           if(s[0] == 'Q')
           {
               ans = 0;
               scanf("%d%d", &l,&r);
               query(1, l, r);
               printf("%lld\n", ans);
           }
           else
           {
               scanf("%d%d%d", &l,&r,&w);
               update(1, l, r, w);
           }
       }
   }

}