POJ3468 A Simple Problem with Integers 题解&代码

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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;}

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