POJ-3468：A Simple Problem with Integers

A Simple Problem with Integers

来源：POJ

标签：数据结构，线段树，区间修改

参考资料：

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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.

输入

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.

输出

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

输入样例

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

输出样例

4
55
9
15

题目大意

解题思路

参考代码

#include
#include
#define MAXN 100005
using namespace std;

struct SegTreeNode{
	long long val;
	long long addMark;
}segTree[MAXN*4];

void build(int root, int arr[], int begin, int end){
	segTree[root].addMark=0;
	if(begin==end){
		segTree[root].val=arr[begin];
	}
	else{
		int mid=(begin+end)/2;
		build(root*2+1,arr,begin,mid);
		build(root*2+2,arr,mid+1,end);
		segTree[root].val=segTree[root*2+1].val+segTree[root*2+2].val;
	}
}

void pushDown(int root,int begin,int end){
	if(segTree[root].addMark!=0){
		segTree[root*2+1].addMark+=segTree[root].addMark;
		segTree[root*2+2].addMark+=segTree[root].addMark;
		
		segTree[root*2+1].val+=((begin+end)/2-begin+1)*segTree[root].addMark;
		segTree[root*2+2].val+=(end-(begin+end)/2)*segTree[root].addMark;
		
		segTree[root].addMark=0;
	}
}

long long query(int root, int begin, int end, int L, int R){
	if(L>end || R<begin) return 0;
	if(L<=begin && R>=end) return segTree[root].val;
	pushDown(root,begin,end);
	int mid=(begin+end)/2;
	return query(root*2+1, begin, mid, L, R)+query(root*2+2, mid+1, end, L, R);
}

void update(int root, int begin, int end, int L, int R, int addVal){
	if(L>end || R<begin){
		return;
	}
	if(L<=begin && R>=end){
		segTree[root].addMark+=addVal;
		segTree[root].val+=(end-begin+1)*addVal;
		return;
	}
	pushDown(root,begin,end);
	int mid=(begin+end)/2;
	update(root*2+1, begin, mid, L, R, addVal);
	update(root*2+2, mid+1, end, L, R, addVal);
	segTree[root].val=segTree[root*2+1].val+segTree[root*2+2].val;
}

int main(){
	int N,Q;
	while(scanf("%d%d",&N,&Q)!=EOF){
		int arr[MAXN];
		for(int i=0;i<N;i++){
			scanf("%d",&arr[i]);
		}
		build(0,arr,0,N-1);
		
		int a,b,c;
		char s[5]; 
		for(int i=0;i<Q;i++){
			scanf("%s",s);
			if(s[0]=='Q'){
				scanf("%d%d",&a,&b);
				printf("%lld\n",query(0,0,N-1,a-1,b-1));
			}
			else{
				scanf("%d%d%d",&a,&b,&c);
				update(0,0,N-1,a-1,b-1,c);
			}
		}
	}
	return 0;
}