A Simple Problem with Integers （基础线段树）

题意
给出了一个序列，你需要处理如下两种询问
“C a b c”表示给[a, b]区间中的值全部增加c (-10000 ≤ c ≤ 10000)。
“Q a b” 询问[a, b]区间中所有值的和。
本题很简单注意数据范围就可以了 （他是很直白的区域更新）lazy 标记要用 long long 在这错了很多次。

以下是AC代码

#include <iostream>#include <stdio.h>using namespace std;const int maxn=100005;struct node{    int l,r;    long long lazy;    long long sum;} tree[maxn*4];int a[maxn+5];void pushdown(int id){    if(tree[id].lazy!=0)    {        tree[id*2].lazy+=tree[id].lazy;        tree[id*2+1].lazy+=tree[id].lazy;        tree[id*2].sum+=(tree[id].lazy*(tree[id*2].r-tree[id*2].l+1));        tree[id*2+1].sum+=(tree[id].lazy*(tree[id*2+1].r-tree[id*2+1].l+1));        tree[id].lazy=0;//注意把lazy标记计零    }}//建树void build (int left,int right,int id)  {    tree[id].l=left;    tree[id].r=right;    tree[id].lazy=0;    int mid=(left+right)/2;    if(tree[id].l==tree[id].r)    {        tree[id].sum=a[tree[id].l];        return ;    }    else    {        build(left,mid,id*2);        build(mid+1,right,id*2+1);    }    tree[id].sum=tree[id*2].sum+tree[id*2+1].sum;}//给给定区间加上一个数void add(int left,int right,int val,int id){    int mid=(tree[id].l+tree[id].r)/2;    if(left<=tree[id].l&&tree[id].r<=right)    {        tree[id].lazy+=val;        tree[id].sum+=(val*(tree[id].r-tree[id].l+1));        return ;    }        //增加的时候也要把lazy标记往下传    pushdown(id);    if(right<=mid)    {        add(left,right,val,id*2);    }    else if(left>=mid+1)    {        add(left,right,val,id*2+1);    }    else    {        add(left,mid,val,id*2);        add(mid+1,right,val,id*2+1);    }    tree[id].sum=tree[id*2].sum+tree[id*2+1].sum;}//查询 long long inquiry(int left,int right,int id){    int mid=(tree[id].l+tree[id].r)/2;    if(left<=tree[id].l&&tree[id].r<=right)    {        return tree[id].sum;    }//注意把lazy标记往下传    pushdown(id);    long long ans=0;    if(right<=mid)    {        ans+=inquiry(left,right,id*2);    }    else if(left>=mid+1)    {        ans+=inquiry(left,right,id*2+1);    }    else    {        ans+=inquiry(left,mid,id*2);        ans+=inquiry(mid+1,right,id*2+1);    }    return ans;}int main (){    int n,m,i;    scanf ("%d %d",&n,&m);    for (i=1; i<=n; i++)        scanf ("%d",&a[i]);    build(1,n,1);    while (m--)    {        char type[2];        int x,y,val;        scanf ("%s",&type);        if(type[0]=='Q')        {            scanf ("%d %d",&x,&y);            long long ans=inquiry(x,y,1);            cout<<ans<<endl;        }        if(type[0]=='C')        {            scanf ("%d %d %d",&x,&y,&val);            add(x,y,val,1);        }    }    return 0;}