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, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a+1, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a+1, ... , A_b.

 

Output

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

 

Sample Input

10 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

 

Sample Output

455915

 

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[n].l = l;    tree[n].r = r;    tree[n].tag = 0;    tree[n].lazy = 0;    if(l == r){        tree[n].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[n].num    tree[n].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[n].l;    int r = tree[n].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[n].lazy = 1;        tree[n].tag += der;        tree[n].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[n].lazy == 1){        tree[n].lazy = 0;        Modify(2*n,l,mid,tree[n].tag);        Modify(2*n+1,mid+1,r,tree[n].tag);        tree[n].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[n].num = tree[2*n].num + tree[2*n+1].num;}long long Find(int n,int x,int y){    int l = tree[n].l;    int r = tree[n].r;    int mid = (l + r) / 2;    if(l == x && r == y){        return tree[n].num;    }    if(tree[n].lazy == 1){        tree[n].lazy = 0;        Modify(2*n,l,mid,tree[n].tag);        Modify(2*n+1,mid+1,r,tree[n].tag);        tree[n].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;}