A Simple Problem with Integers//POJ - 3468//线段树

A Simple Problem with Integers//POJ - 3468//线段树

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题目

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.
题意
求区间和，中间会有改变
链接https://vjudge.net/contest/352439#problem/B

思路

线段树模板题

代码

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
using namespace std;
typedef long long ll;
ll n,m;
int a[100100],val;
void pushdown(int node);

struct seg{
	ll sum,lazy;
	int l,r;
}segtree[100100*4];

void build(int node,int l,int r){
    segtree[node].l=l;
	segtree[node].r=r;
	segtree[node].lazy=0;
    if(l==r){
    	segtree[node].sum=a[l];
		return;
	}
	else{
		int mid=(l+r)>>1;
	    build(node<<1,l,mid);
	    build(node<<1|1,mid+1,r);
	    segtree[node].sum = segtree[node<<1].sum + segtree[node<<1|1].sum;
	}
}

ll query(int node,int x,int y){         //求和
	if(segtree[node].l == x && segtree[node].r == y){
		return segtree[node].sum;
	}
	pushdown(node);
	int mid = segtree[node].l + segtree[node].r;
	ll res = 0;
	mid >>= 1;
	if(mid >= y)
		res += query(node<<1,x,y);
	else if(mid < x)
		res += query(node<<1|1,x,y);
	else{
		res += query(node<<1,x,mid);
		res += query(node<<1|1,mid+1,y);
	}
	return res;
}

void pushdown(int node){
	if(segtree[node].lazy != 0){
		segtree[node<<1].lazy += segtree[node].lazy;
		segtree[node<<1|1].lazy += segtree[node].lazy;
		segtree[node<<1].sum += segtree[node].lazy * (segtree[node<<1].r - segtree[node<<1].l + 1);
		segtree[node<<1|1].sum += segtree[node].lazy * (segtree[node<<1|1].r - segtree[node<<1|1].l + 1);;
		segtree[node].lazy = 0;
	}
}

void update(int node,int x,int y){					//更新
	if(segtree[node].l == x && segtree[node].r == y){
		segtree[node].lazy += val;
		segtree[node].sum += val * (y - x + 1);
		return;
	}
	if(segtree[node].l == segtree[node].r)
		return;
	pushdown(node);
	int mid = segtree[node].l + segtree[node].r;
	mid >>= 1;
	if(mid >= y)
		update(node<<1,x,y);
	else if(mid < x)
		update(node<<1|1,x,y);
	else{
		update(node<<1,x,mid);
		update(node<<1|1,mid+1,y);
	}
	segtree[node].sum = segtree[node<<1].sum + segtree[node<<1|1].sum;
}

int main()
{
    char ch[5];
    scanf("%lld%lld",&n,&m);
    for(ll i=1;i<=n;i++)
        scanf("%lld",&a[i]);
    build(1,1,n);
    for(ll i=1;i<=m;i++){
        scanf("%s",ch);
        if(ch[0]=='C'){
            ll x,y;
            scanf("%lld%lld%lld",&x,&y,&val);
            update(1,x,y);
        }
        else{
            ll x,y;
            scanf("%lld%lld",&x,&y);
            ll ans=query(1,x,y);
            printf("%lld\n",ans);
        }
    }
    return 0;
}

注意