poj 3468 A Simple Problem with Integers LAZY线段树

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K

Case Time Limit: 2000MS

Description

You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.

Input

The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.

Output

You need to answer all Q commands in order. One answer in a line.

Sample Input

10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output

4
55
9
15

思路：基础的懒散线段树练习   (水完这道就不水了)

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<vector>
#include <ctime>
#include<queue>
#include<set>
#include<map>
#include<stack>
#include<iomanip>
#include<cmath>
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define maxn 0x3f3f3f3f
#define MAX 1000100
///#pragma comment(linker, "/STACK:102400000,102400000")
typedef long long ll;
typedef unsigned long long ull;
#define INF (1ll<<60)-1
using namespace std;
int n,q;
ll a[400100],sum[400100],lazy[400100];
char s[10];
void build(int root,int l,int r){
    if(l==r){
        sum[root]=a[l];
        return ;
    }
    int mid=(l+r)/2;
    build(root*2,l,mid);
    build(root*2+1,mid+1,r);
    sum[root]=sum[root*2]+sum[root*2+1];
}
void pushdown(int root,int l){
    if(lazy[root]){
        lazy[root*2]+=lazy[root];
        lazy[root*2+1]+=lazy[root];
        sum[root*2]+=(ll)(l-l/2)*lazy[root];
        sum[root*2+1]+=(ll)(l/2)*lazy[root];
        lazy[root]=0;
    }
}
void update(int root,int l,int r,int x,int y,ll z){
    if(l>=x && r<=y){
        sum[root]+=(ll)(r-l+1)*z;
        lazy[root]+=z;
        return ;
    }
    pushdown(root,r-l+1);
    int mid=(l+r)/2;
    if(y<=mid) update(root*2,l,mid,x,y,z);
    else if(x>mid) update(root*2+1,mid+1,r,x,y,z);
    else {
        update(root*2,l,mid,x,mid,z);
        update(root*2+1,mid+1,r,mid+1,y,z);
    }
    sum[root]=sum[root*2]+sum[root*2+1];
}
ll query(int root,int l,int r,int x,int y){
    if(l>=x && r<=y){
        ///cout<<"root="<<root<<" "<<sum[root]<<endl;
        return sum[root];
    }
    pushdown(root,r-l+1);
    int mid=(l+r)/2;
    if(y<=mid) return query(root*2,l,mid,x,y);
    else if(x>mid) return query(root*2+1,mid+1,r,x,y);
    else {
        return query(root*2,l,mid,x,mid)+query(root*2+1,mid+1,r,mid+1,y);
    }
}
int main(){
    while(scanf("%d%d",&n,&q)!=EOF){
        mst(sum,0);
        mst(lazy,0);
        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
        build(1,1,n);
        while(q--){
            scanf("%s",s);
            int x,y;
            ll z;
            if(s[0]=='Q'){
                scanf("%d%d",&x,&y);
                printf("%lld\n",query(1,1,n,x,y));
            } else {
                scanf("%d%d%lld",&x,&y,&z);
                update(1,1,n,x,y,z);
            }
        }
    }
    return 0;
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