A Simple Problem with Integers(线段树，区间更新)

整数问题与线段树

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 82 阅读

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#数据结构与算法

本文介绍了一个涉及整数操作的问题及解决方案。通过使用线段树数据结构，有效地处理了区间加法与区间求和两种操作。文章提供了完整的代码实现，并详细解释了关键部分，如区间更新和查询。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 83822		Accepted: 25942
Case Time Limit: 2000MS

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 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

Hint

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

Source

题解：线段树区间更新模版，由于mid比较那块出问题了，错了好半天。。。还有就是注意左边的要是x-（x>>1），因为左边可能多一

代码：

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define SL(x) scanf("%lld",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define PL(x) printf("%lld",x)
typedef long long LL;
const int INF=0x3f3f3f3f;
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
#define S(x) tree[x].sum
#define L(x) tree[x].lazy
const int MAXN=100010;
LL ans=0;
struct Node{
	LL lazy,sum;
};
Node tree[MAXN<<2];

void pushup(int root){
	S(root)=S(ll)+S(rr);
}
void pushdown(int root,int x){
	if(L(root)){
		L(ll)+=L(root);
		L(rr)+=L(root);
		S(ll)+=L(root)*(x-(x>>1));//注意 
		S(rr)+=L(root)*(x>>1);
		L(root)=0;
	}
}
void build(int root,int l,int r){
	int mid=(l+r)>>1;
	L(root)=0;
	if(l==r){
		SL(S(root));
		//printf("%d %d %lld\n",l,r,S(root));
		return;
	}
	build(lson);
	build(rson);
	pushup(root);
}
void update(int root,int l,int r,int A,int B,int v){
		int mid=(l+r)>>1;
	if(l>=A&&r<=B){
		L(root)+=v;//注意是+= 
		S(root)+=v*(r-l+1);
		return;
	}
	pushdown(root,r-l+1);
	if(mid>=A)update(lson,A,B,v);//注意 
	if(mid<B)update(rson,A,B,v);//注意 
	pushup(root);
}
void query(int root,int l,int r,int A,int B){
		int mid=(l+r)>>1;
		//printf("%d %d %lld\n",l,r,S(root));
	//	PI(A);P_;PI(B);
	if(l>=A&&r<=B){
		ans+=S(root);
		//printf("%d %d %lld\n",l,r,S(root));
		return;
	}
	pushdown(root,r-l+1);
	if(mid>=A)query(lson,A,B);//
	if(mid<B)query(rson,A,B);//
}
int main(){
	int N,Q;
	char s[2];
	int A,B,v;
	while(~scanf("%d%d",&N,&Q)){
		build(1,1,N);
		while(Q--){
			scanf("%s",s);
			if(s[0]=='Q'){
				scanf("%d%d",&A,&B);
				ans=0;
				query(1,1,N,A,B);
				printf("%lld\n",ans);
			}
			else{
				scanf("%d%d%d",&A,&B,&v);
				update(1,1,N,A,B,v);
			}
		}
	}
	return 0;
}

extern "C++"{
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace  std;
typedef long long LL;
typedef unsigned u;
typedef unsigned long long ull;
#define mem(x,y) memset(x,y,sizeof(x))
void SI(double &x){scanf("%lf",&x);}
void SI(int &x){scanf("%d",&x);}
void SI(LL &x){scanf("%lld",&x);}
void SI(u &x){scanf("%u",&x);}
void SI(ull &x){scanf("%llu",&x);}
void SI(char *s){scanf("%s",s);}

void PI(int x){printf("%d",x);}
void PI(double x){printf("%lf",x);}
void PI(LL x){printf("%lld",x);}
void PI(u x){printf("%u",x);}
void PI(ull x){printf("%llu",x);}
void PI(char *s){printf("%s",s);}

#define NL puts("");
#define ll root<<1
#define rr root<<1|1
#define lson ll,l,mid
#define rson rr,mid+1,r
const int INF=0x3f3f3f3f;
const int MAXN=100010;
LL tree[MAXN << 2];
LL lazy[MAXN << 2];
}

void pushup(int root){
    tree[root] = tree[ll] + tree[rr];
}
void pushdown(int root,int x){
    if(lazy[root]){
        lazy[ll] += lazy[root];
        lazy[rr] += lazy[root];
    //    tree[ll]=lazy[root]*(mid-l+1);
    //    tree[rr]=lazy[root]*(r-mid);
        tree[ll] += lazy[root] * (x - (x >> 1));
        tree[rr] += lazy[root] * (x >> 1);
        
        lazy[root] = 0;
    }
}
void build(int root,int l,int r){
    int mid = (l + r) >> 1;
    lazy[root] = 0;
    if(l == r){
        SI(tree[root]);
        return ;
    }
    build(lson);
    build(rson);
    pushup(root);
}
void update(int root,int l,int r,int L,int R,int C){
    if(l >= L && r <= R){
        tree[root] += (r - l + 1) * C;
        lazy[root] += C;
        return ;
    }
    int mid = (l + r) >> 1;
    pushdown(root,r-l+1);
    if(mid >= L)
        update(lson,L,R,C);
    if(mid < R)
        update(rson,L,R,C);
    pushup(root);
}
LL query(int root,int l,int r,int L,int R){
    int mid = (l + r) >> 1;
    if(l >= L && r <= R){
        return tree[root];
    }
    pushdown(root,r-l+1);
    LL ans = 0;
    if(mid >= L)ans += query(lson,L,R);
    if(mid < R)ans += query(rson,L,R);
    return ans;
}
int main(){
    //assert(false);
    int N,M;
    while(~scanf("%d%d",&N,&M)){
        build(1,1,N);
        char s[5];
        int l,r,c;
        while(M--){
            SI(s);
            SI(l);SI(r);
            if(s[0]=='Q'){
                PI(query(1,1,N,l,r)),NL
            }
            else{
                SI(c);
                update(1,1,N,l,r,c);
            }
        }
    }
    return 0;
}