POJ3468 A Simple Problem with Integers 题解&代码

题意：给出n个数排成一列，有q个操作，分为Q和C，Q操作[l,r]询问区间[l,r]的区间和，C操作[l,r]对[l,r]区间同时增加x，按题意输出就行了
题解：线段树，主要是好久没写又错了好长一段时间…万年LL我就不多说了，看错范围觉得sum[]不需要LL，另外重点是query和insert不可能越界访问，所以就算区间对于这组数据是无效区间，只要访问到对应区间也一定要pushdown…忘了结果纠结了好久，小数据还找不出问题

#include<iostream>
#include<cstdio>
#define LL long long
#define lson (o<<1)
#define rson ((o<<1)|1)
using namespace std;
const int maxn = 100005;
int n,q,l,r,c,a[maxn],lazy[4*maxn];
LL sum[4*maxn];
char s[5];
void build(int o,int l,int r)
{
    if(l==r)
    {
        sum[o]=(LL)a[l];
        return;
    }
    int mid = (l+r)/2;
    build(lson,l,mid);
    build(rson,mid+1,r);
    sum[o]=sum[lson]+sum[rson];
}
void pushdown(int o,int l,int r)
{
    if(lazy[o] && l!=r)
    {
        int mid = (l+r)/2;
        lazy[lson]+=lazy[o];
        lazy[rson]+=lazy[o];
        sum[lson]+=(LL)lazy[o]*(LL)(mid-l+1);
        sum[rson]+=(LL)lazy[o]*(LL)(r-mid);
    }
    lazy[o]=0;
}
void Insert(int o,int l,int r,int L,int R,int c)
{
    pushdown(o,l,r);
    if(l>=L && r<=R)
    {
        lazy[o]+=c;
        sum[o]+=(LL)(r-l+1)*(LL)lazy[o];
        pushdown(o,l,r);
        return;
    }
    if(l>R || r<L)return;
    int mid = (l+r)/2;
    Insert(lson,l,mid,L,R,c);
    Insert(rson,mid+1,r,L,R,c);
    sum[o]=sum[lson]+sum[rson];
}
LL query(int o,int l,int r,int L,int R)
{
    pushdown(o,l,r);
    if(l>=L && r<=R)return sum[o];
    if(l>R || r<L)return 0;
    int mid = (l+r)/2;
    LL ret=query(lson,l,mid,L,R)+query(rson,mid+1,r,L,R);
    sum[o]=sum[lson]+sum[rson];
    return ret;
}
int main(void)
{
    scanf("%d%d",&n,&q);
    for(int i = 1; i <= n; i++)
        scanf("%d",&a[i]);
    build(1,1,n);
    for(int i = 0; i < q; i++)
    {
        scanf("%s%d%d",s,&l,&r);
        if(s[0]=='C')
        {
            scanf("%d",&c);
            Insert(1,1,n,l,r,c);
        }
        else printf("%lld\n",query(1,1,n,l,r));
    }
    return 0;
}