A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 138575		Accepted: 42915
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

线段树练习、AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
#define lson rt<<1,left,m
#define rson rt<<1|1,m+1,right 
const int maxn = 1e5 + 10;
ll sum[maxn<<2],add[maxn<<2]; // sum区间和 add延迟标记数组
struct node{
	int left,right;
	int mid(){
		return (left+right)>>1;
	}
}tree[maxn<<2]; 
void PushUp(int rt){ //向上更新 
	sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}
void PushDown(int rt,int m){
	if(add[rt]){
		add[rt<<1] += add[rt];
		add[rt<<1|1] += add[rt];
		sum[rt<<1] += add[rt]*(m-(m>>1));
		sum[rt<<1|1] += add[rt]*(m>>1);
		add[rt] = 0;
	}
}
void BuildTree(int rt,int left,int right){
	tree[rt].left = left;
	tree[rt].right = right;
	add[rt] = 0;
	if( left == right){
		scanf("%lld",&sum[rt]);
		return ;
	}
	int m = tree[rt].mid();
	BuildTree(lson);
	BuildTree(rson);
	PushUp(rt);
}

void Update(int c,int left,int right,int rt){
	if(tree[rt].left == left && tree[rt].right == right){
		add[rt] += c;
		sum[rt] += (ll)c*(right-left+1);
		return;
	}
	if(tree[rt].left == tree[rt].right ) return ;
	PushDown(rt,tree[rt].right-tree[rt].left+1);
	int m = tree[rt].mid();
	if(right <= m) Update(c,left,right,rt<<1);
	else if(left > m ) Update(c,left,right,rt<<1|1);
	else{
		Update(c,left,m,rt<<1);
		Update(c,m+1,right,rt<<1|1);
	}	
	PushUp(rt);
}

ll Query(int rt,int left,int right){
	if(tree[rt].left == left && tree[rt].right == right){
		return sum[rt];
	}
	PushDown(rt,tree[rt].right-tree[rt].left+1);
	int m = tree[rt].mid();
	ll ans = 0;
	if(right <= m ) ans+=Query(rt<<1,left,right);
	else if(left > m) ans+=Query(rt<<1|1,left,right);
	else{
		ans+=Query(rt<<1,left,m);
		ans+=Query(rt<<1|1,m+1,right);
	}
	return ans;
}
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m)){
		BuildTree(1,1,n);	
		char ch[2];
		int a,b,c;
		while(m--){
			scanf("%s",ch);
			if(ch[0] == 'Q'){
				scanf("%d%d",&a,&b);
				printf("%lld\n",Query(1,a,b));
			}
			else{
				scanf("%d%d%d",&a,&b,&c);
				Update(c,a,b,1);
			}
		}
	}
	return 0;
}