线段树 P3372【区间修改 区间查询】

题目

https://www.luogu.com.cn/problem/P3372

代码

树状数组方法

#include<iostream>
#include<cstdio>
using namespace std;
long long a[100001], c[100001],sum[100001];
int n, m;
long long  lowbit(long long x)
{
    return x&(-x);
}
void update(long long i, long long k)
{
    long long x = i;
    while (i <= n)
    {
        c[i] += k;
        sum[i] += (x-1)*k;
        i += lowbit(i);
    }
}
long long getsum(long long i)
{
    long long res = 0;
    long long x = i;
    while (i > 0)
    {
        res += c[i]*x-sum[i];
        i -= lowbit(i);
    }
    return res;
}
int main()
{
    long long type,x,y,k;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        update(i, a[i]-a[i-1]);
    }
    for (int i = 0; i < m; i++)
    {
        scanf("%lld %lld %lld", &type,&x,&y);
        if (type == 1)
        {
            scanf("%lld", &k);
            update(x, k);
            update(y + 1, -k);
        }
        else
        {
            printf("%lld
", getsum(y) - getsum(x - 1));
        }
    }
}

开始int 溢出，改为long long通过

 线段树方法

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
#define MAX 100001
#define INF 0x3f3f3f3f
long long a[MAX], c[MAX << 2], lazy[MAX << 2];//空间，一般会开到4*n的空间防止RE。
int n, m;
void build(long long k, int l, int r)
{
    if (l == r)//如果是最后的叶子节点，就直接赋值
        c[k] = a[l];
    else
    {
        int m = l + ((r - l) >> 1);//注意这里的括号，进行优先级的处理
        build(k << 1, l, m);//不是的话就递归
        build(k << 1 | 1, m + 1, r);
        c[k] = c[k << 1] + c[k << 1 | 1];//由于不是最后的叶子节点，所以一定有左右子节点，所以此节点的值就是左右子节点较大的值
    }
}

void pushdown(long long k, int l, int r)
{
    if (lazy[k])
    {
        int m = (l + r) >> 1;
        lazy[k << 1] += lazy[k];
        lazy[k << 1 | 1] += lazy[k];
        c[k << 1] += lazy[k] * (m - l + 1);
        c[k << 1 | 1] += lazy[k] * (r - (m + 1) + 1);
        lazy[k] = 0;
    }
}
void update(int L, int R, int v, int l, int r, long long k)
{
    if (L <= l&&R >= r)
    {
        c[k] += (r - l + 1)* v;
        lazy[k] += v;
    }
    else
    {
        pushdown(k, l, r);
        int m = l + ((r - l) >> 1);
        if (L <= m)
            update(L, R, v, l, m, k << 1);
        if (R >m)
            update(L, R, v, m + 1, r, k << 1 | 1);
        c[k] = c[k << 1] + c[k << 1 | 1];
    }

}
long long query(int L, int R, int l, int r, long long k)
{
    if (L <= l&&R >= r)
        return c[k];
    else
    {
        pushdown(k, l, r); /**每次都需要更新子树的Lazy标记*/
        long long res = 0;
        int m = l + ((r - l) >> 1);
        if (L <= m)res += query(L, R, l, m, k << 1);
        if (m < R)res += query(L, R, m + 1, r, k << 1 | 1);
        return res;
    }
}

int main()
{
        long long type,x,y,k;
        scanf("%d %d", &n, &m);
        for (int i = 1; i <= n; i++)
            scanf("%lld", &a[i]);
        build(1, 1, n);
        for (int i = 0; i < m; i++)
        {
            scanf("%lld %lld %lld", &type,&x,&y);
            if (type == 1)
            {
                scanf("%lld", &k);
                update(x, y, k, 1, n, 1);
            }
            else
            {
                printf("%lld
", query(x,y,1,n,1));
            }
        }
}