POJ3468：A Simple Problem with Integers——题解

http://poj.org/problem?id=3468

实现一个线段树，能够做到区间修改和区间查询和。

明显板子题。

#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<iostream>
using namespace std;
typedef long long ll;
inline ll read(){
    ll X=,w=; char ch=;
    while(ch<'' || ch>'') {w|=ch=='-';ch=getchar();}
    while(ch>='' && ch<='') X=(X<<)+(X<<)+(ch^),ch=getchar();
    return w?-X:X;
}
ll tree[],b[];
ll lazy[];
void build(int a,int l,int r){
    lazy[a]=;
    if(l==r){
    tree[a]=b[l];
    return;
    }
    int mid=(l+r)/;
    build(a*,l,mid);
    build(a*+,mid+,r);
    tree[a]=tree[a*]+tree[a*+];
    return;
}
inline void push(int a,int l,int mid,int r){
    lazy[a*+]+=lazy[a];
    lazy[a*]+=lazy[a];
    tree[a*+]+=lazy[a]*(r-mid);
    tree[a*]+=lazy[a]*(mid-l+);
    lazy[a]=;
    return;
}
ll wen(int a,int l,int r,int l1,int r1){
    if(l1<=l&&r1>=r){
    return tree[a];
    }
    if (l1>r||r1<l) return ;
    int mid=(l+r)/;
    push(a,l,mid,r);
    return wen(a*,l,mid,l1,r1)+wen(a*+,mid+,r,l1,r1);
}
void gai(int a,int l,int r,int l1,int r1,int add){
    if (l1>r||r1<l) return;
    if(l1<=l&&r1>=r){
    lazy[a]+=add;
    tree[a]+=add*(r-l+);
    return;
    }
    int mid=(l+r)/;
    push(a,l,mid,r);
    gai(a*,l,mid,l1,r1,add);
    gai(a*+,mid+,r,l1,r1,add);
    tree[a]=tree[a*]+tree[a*+];
    return;
}
int main(){
    int m=read();
    int ha=read();
    for(int i=;i<=m;i++){
    b[i]=read();
    }
    build(,,m);
    for(int i=;i<=ha;i++){
    char w;
    cin>>w;
    if(w=='C'){
        int a1=read();
        int a2=read();
        int a3=read();
        gai(,,m,a1,a2,a3);
    }else{
        int x=read();
        int y=read();
        printf("%lld\n",wen(,,m,x,y));
    }
    }
    return ;
}