Poj A Simple Problem with Integers（lazy线段树）

A Simple Problem with Integers

Time Limit : 10000/5000ms (Java/Other) Memory Limit : 262144/131072K (Java/Other)

Total Submission(s) : 40 Accepted Submission(s) : 13

Problem 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

Source

PKU

注意要用longlong。。。

#include <stdio.h>
#include <algorithm>
using namespace std;

struct Node{
    int l,r;
    int lazy;
    long long num,tag;
}tree[1000000];
long long A[100005];

void Build(int n,int l,int r){
    //先将一棵树的各个元素初始化
    tree
.l = l;
    tree
.r = r;
    tree
.tag = 0;
    tree
.lazy = 0;
    if(l == r){
        tree
.num = A[l];
        return;
    }
    int mid = (l + r) / 2;
    Build(2*n,l,mid);
    Build(2*n+1,mid+1,r);
    //分别建好一棵树的左右儿子之后，计算tree
.num
    tree
.num = tree[2*n].num + tree[2*n+1].num;
}

//lazy标记的思想是修改到到确切的区间后，不再修改儿子
//只是用一个lazy标记，等以后修改或查找到这个区间
//再把lazy标记往下传
void Modify(int n,int x,int y,long long der){
    //先得到所在区间的范围
    int l = tree
.l;
    int r = tree
.r;
    int mid = (l + r) / 2;
    //找到恰好的区间进行修改及lazy标记
    if(l == x && r == y){
        tree
.lazy = 1;
        tree
.tag += der;
        tree
.num += der * (y - x + 1);
        return;
    }
    //如果该区间有lazy标记，则修改儿子，并将标记清0
    if(tree
.lazy == 1){
        tree
.lazy = 0;
        Modify(2*n,l,mid,tree
.tag);
        Modify(2*n+1,mid+1,r,tree
.tag);
        tree
.tag = 0;
    }
    if(x <= mid)//修改左儿子的充要条件
        Modify(2*n,x,min(y,mid),der);
    if(y > mid)//修改右儿子的充要条件
        Modify(2*n+1,max(mid+1,x),y,der);
    tree
.num = tree[2*n].num + tree[2*n+1].num;
}

long long Find(int n,int x,int y){
    int l = tree
.l;
    int r = tree
.r;
    int mid = (l + r) / 2;
    if(l == x && r == y){
        return tree
.num;
    }
    if(tree
.lazy == 1){
        tree
.lazy = 0;
        Modify(2*n,l,mid,tree
.tag);
        Modify(2*n+1,mid+1,r,tree
.tag);
        tree
.tag = 0;
    }
    long long ans = 0;
    if(x <= mid)    ans += Find(2*n,x,min(mid,y));
    if(y > mid)     ans += Find(2*n+1,max(mid+1,x),y);
    return ans;
}

int main(){
    int n,q;
    int i;
    char op;
    int a,b,c;

    while(scanf("%d%d",&n,&q) != EOF){
        for(i = 1;i <= n;i++){
            scanf("%lld",&A[i]);
        }
        Build(1,1,n);
        for(i = 0;i < q;i++){
            getchar();
            scanf("%c",&op);
            if(op == 'Q'){
                scanf("%d%d",&a,&b);
                printf("%lld\n",Find(1,a,b));
            }
            else{
                scanf("%d%d%d",&a,&b,&c);
                Modify(1,a,b,c);
            }
        }
    }
    return 0;
}