poj3468 A simple problem with integers

线段树区间更新与查询

最新推荐文章于 2023-10-02 10:26:46 发布

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本文介绍了一种使用线段树进行区间更新和查询的高效算法，通过实例讲解了如何实现线段树的构建、更新和查询操作，解决了大规模数据集上的区间加法和求和问题。

You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

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.

简单的线段树+区间更新模板看到最后hint已经把类型都改为longlong了

但是还是WA了一发发现输入的时候用的%d 改成%I64d 就AC了


#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define inf 1e18
using namespace std;

int q, n;
const int maxn = 100005;
long long val[maxn], tree[maxn << 2],lazy[maxn << 2];


void pushup(int rt)//更新
{
    tree[rt] = tree[rt << 1] + tree[rt << 1 | 1];
}

void build(int l, int r, int rt)
{
    if(l == r){
        tree[rt] = val[l];
        return;
    }
    int m = (l + r) >> 1;
    build(l, m, rt << 1);
    build(m + 1, r, rt << 1 | 1);
    pushup(rt);
}

void pushdown(int rt, int ln, int rn)
{
    if(lazy[rt]){
        lazy[rt << 1] += lazy[rt];
        lazy[rt << 1 | 1] += lazy[rt];
        tree[rt << 1] += lazy[rt] * ln;
        tree[rt << 1 | 1] += lazy[rt] * rn;
        lazy[rt] = 0;
    }
}

void update(int L, int C, int l, int r, int rt)
{
    if(l == r){
        tree[rt] += C;
        return;
    }
    int m = (l + r) >>1;
    if(L <= m) update(L, C, l, m, rt << 1);
    else update(L, C, m + 1, r, rt << 1 | 1);
    pushup(rt);
}

void update(int L, int R, int C, int l, int r, int rt)
{
    if(L <= l && r <= R){//本区间完全在操作区间内
        tree[rt] += C * (r - l + 1);
        lazy[rt] += C;
        return;
    }
    int m = (l + r) >> 1;
    pushdown(rt, m - l + 1, r - m);
    if(L <= m) update(L, R, C, l, m, rt << 1);
    if(R > m) update(L, R, C, m + 1, r, rt << 1 | 1);
    pushup(rt);
}

long long query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R){
        return tree[rt];
    }
    int m = (l + r) >> 1;
    pushdown(rt, m - l + 1, r - m);

    long long ans = 0;
    if(L <= m) ans += query(L, R, l, m, rt << 1);
    if(R > m) ans += query(L, R, m + 1, r, rt << 1 | 1);
    return ans;
}

int main()
{
    while(scanf("%d%d", &n, &q)!= EOF){
        for(int i = 1; i <= n; i++){
            scanf("%I64d", &val[i]);
        }
        build(1, n, 1);

        while(q--){
            getchar();
            char ch;
            int a, b, c;
            scanf("%c", &ch);
            if(ch == 'Q'){
                scanf("%d%d", &a, &b);
                cout<<query(a, b, 1, n, 1)<<endl;
            }
            else{
                scanf("%d%d%d", &a, &b, &c);
                update(a, b, c, 1, n, 1);
            }
        }
    }
    return 0;
}