POJ_3468 A Simple Problem with Integers 【线段树区间查询+修改】

一、题目

POJ3468

二、分析

裸的线段树区间查询+修改。

三、AC代码

#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;

#define ll long long
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
const int MAXN = 1e5 + 3;
ll add[MAXN << 2], sum[MAXN << 2];

void PushUp(int rt)
{
    sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    return;
}

void PushDown(int ln, int rn, int rt)
{
    if(add[rt])
    {
        sum[rt << 1] += (ll) ln * add[rt];
        sum[rt << 1 | 1] += (ll) rn * add[rt];
        add[rt << 1] += add[rt];
        add[rt << 1 | 1] += add[rt];
        add[rt] = 0;
    }
    return;
}

void Build(int l, int r, int rt)
{
    add[rt] = 0;
    if(l == r)
    {
        scanf("%lld", &sum[rt]);
        return;
    }
    int mid = (l + r) >> 1;
    Build(lson);
    Build(rson);
    PushUp(rt);
    return;
}

void Update(int L, int R, int c, int l, int r, int rt)
{
    if(L <= l && r <= R)
    {
        add[rt] += c;
        sum[rt] += (ll)(r - l + 1) * c;
        return;
    }
    int mid = (l + r) >> 1;
    PushDown(mid - l + 1, r - mid, rt);
    if(L <= mid)
    {
        Update(L, R, c, lson);
    }
    if(R > mid)
    {
        Update(L, R, c, rson);
    }
    PushUp(rt);
    return;
}

ll Query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R)
        return sum[rt];
    int mid = (l + r) >> 1;
    PushDown(mid - l + 1, r - mid, rt);
    ll ans = 0;
    if(L <= mid)
    {
        ans += Query(L, R, lson);
    }
    if(R > mid)
    {
        ans += Query(L, R, rson);
    }
    return ans;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int N, Q;
    scanf("%d %d", &N, &Q);
    Build(1, N, 1);
    char s[2];
    for(int i = 0; i < Q; i++)
    {
        int L, R, C;
        scanf("%s", s);
        if(s[0] == 'Q')
        {
            scanf("%d %d", &L, &R);
            printf("%I64d
", Query(L, R, 1, N, 1));
        }
        else
        {
            scanf("%d %d %d", &L, &R, &C);
            Update(L, R, C, 1, N, 1);
        }
    }
}