Poj 3468 A Simple Problem with Integers（线段树_the second line contains n integers, where the i-t-优快云博客

本文链接：https://blog.youkuaiyun.com/wang2332/article/details/70170674

本文介绍了一道关于整数操作的问题，并提供了一个使用线段树数据结构解决该问题的C++代码示例。通过维护区间求和及懒惰传播，实现了高效地更新和查询区间内的整数总和。

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A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 107094		Accepted: 33429
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

4 

55 

9 

15 

线段树模板题（虽然我还是不会写 

首先数据量比较大 要用LL 

直接套模板 修改单个点会T不动脑子 

看了看dalao的博客 改了改代码 

解析在代码内2333

#include <cstdio>
#include <algorithm>
using namespace std;
#define LL long long
const int N = 1000000*4+10;
int arr[N];
struct node {
    int left, right;
    LL sum, num;
}tree[N];
void build(int l,int r,int step)
{
    tree[step].left = l;
    tree[step].right = r;
    tree[step].num = 0;
    if(l == r) {
        tree[step].sum = arr[l];
        return ;    
    }
    int mid = (l+r) >> 1;
    build(l,mid,step<<1);
    build(mid+1,r,step<<1|1);
    tree[step].sum = tree[step<<1].sum + tree[step<<1|1].sum;
}
void update(int l,int r,LL value,int step)
{
    if(tree[step].left==l&&r==tree[step].right) {
        tree[step].num += value;
        return;
    }
    tree[step].sum += value*(r-l+1); 
    int mid = (tree[step].left+tree[step].right) >> 1;
    if(r <= mid) update(l,r,value,step<<1);
    else if(l > mid) update(l,r,value,step<<1|1);
    else {
        update(l,mid,value,step<<1);
        update(mid+1,r,value,step<<1|1);
    }
}

LL query(int l,int r,int step)
{
    if(l==tree[step].left && r==tree[step].right) {
        return  tree[step].sum+(r-l+1)*tree[step].num;
    }
    int mid = (tree[step].left+tree[step].right) >> 1;
    tree[step].sum += (tree[step].right-tree[step].left+1)*tree[step].num;
    update(tree[step].left,mid,tree[step].num,step<<1);
    update(mid+1,tree[step].right,tree[step].num,step<<1|1);
    tree[step].num = 0;
    if(r <= mid) return query(l,r,step<<1);
    else if(l > mid) return query(l,r,step<<1|1);
    else return query(l,mid,step<<1)+query(mid+1,r,step<<1|1); 
}

int main()
{
    int n, q, x;
    scanf("%d%d",&n,&q);
    for(int i = 1;i <= n; i++) {
        scanf("%d",&arr[i]);
    }
    build(1,n,1);
    while(q--) {
        getchar();
        char ch;
        int u, v, z; 
        scanf("%c",&ch);
        if(ch == 'C') {
            scanf("%d%d%d",&u,&v,&z);
            update(u,v,z,1);
        }
        if(ch == 'Q') {
            scanf("%d%d",&u,&v);
            printf("%lld\n",query(u,v,1));
        }
    }
return 0;
}