POJ - 3468 A Simple Problem with Integers (线段树 区间更新 增加值）

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 137519		Accepted: 42602
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

Hint

The sums may exceed the range of 32-bit integers.

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Source

POJ Monthly--2007.11.25, Yang Yi

线段树板子题，区间[a,b]内的每个数都加ans；

#include <iostream>//区间更新；注意和为long long
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
#define maxn 100010
ll n,q;
ll tree[maxn<<2], vis[maxn<<2]={0};//用vis数组做延迟标记

void build(ll l,ll r,ll rt){
	if(l == r){
		scanf("%lld",&tree[rt]);
		return;
	}
	ll mid=(l+r)>>1;
	build(l,mid,rt<<1);
	build(mid+1,r,rt<<1|1);
	tree[rt] = tree[rt<<1]+tree[rt<<1|1];
}

void solve(ll l,ll r,ll rt){//用来处理vis数组
	if(vis[rt] == 0)  return;//没有标记证明这个区间没有要加的数
	ll mid=(l+r)>>1;
	tree[rt<<1] += vis[rt]*(mid+1-l);//左孩子
	tree[rt<<1|1] += vis[rt]*(r-mid);//右孩子
	vis[rt<<1] += vis[rt];//标记往下一层，左右孩子都要加；
	vis[rt<<1|1] += vis[rt];
	vis[rt] = 0;//注意清零
}

void update(ll a,ll b,ll ans,ll l,ll r,ll rt){//区间[a,b]内的每个数都加ans
	if(a <= l && b >= r){
		tree[rt] += ans*(r+1-l);//区间有几个数就加几个ans
		vis[rt] += ans;//vis数组标记，每个子节点都加上ans
		return;
	}
	solve(l,r,rt);
	ll mid=(l+r)>>1;
	if(a <= mid)  update(a,b,ans,l,mid,rt<<1);
	if(b > mid)   update(a,b,ans,mid+1,r,rt<<1|1);
	tree[rt] = tree[rt<<1]+tree[rt<<1|1];
}

ll ask(ll a,ll b,ll l,ll r,ll rt){
	if(a <= l && b >= r){
		return tree[rt];
	}
	solve(l,r,rt);//注意查询的时候也要处理一下vis数组
	ll mid=(l+r)>>1;
	ll ans = 0;
	if(a <= mid)  ans += ask(a,b,l,mid,rt<<1);
	if(b > mid)   ans += ask(a,b,mid+1,r,rt<<1|1);
	return ans;
}

int main(){
	scanf("%d%d",&n,&q);
	build(1,n,1);
	char c;
	ll a,b,x;
	while(q--){
		cin>>c;
		if(c == 'Q') {
			scanf("%lld%lld",&a,&b);
			printf("%lld\n",ask(a,b,1,n,1));
		}
		else  {
			scanf("%lld %lld %lld",&a,&b,&x);
			update(a,b,x,1,n,1);
		}
	}
    return 0;
}