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| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 116721 | Accepted: 36266 | |
| 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
Source
用到了指针->,感觉好难
Select Code
#include<iostream>
#include<cstdio>
using namespace std;
struct CNode
{int L,R;CNode *pLeft, *pRight;long long nSum; //原来的和long long lnc; //增量C的累加} ;CNode Tree[200010]; //二倍叶子节点的数目就够了int nCount=0;int Mid(CNode *pRoot){return (pRoot->L + pRoot->R )/2;}void BuildTree(CNode *pRoot, int L, int R){pRoot->L = L;pRoot->R = R;pRoot->nSum = 0;pRoot->lnc = 0;if(L==R) return ;nCount++;pRoot->pLeft = Tree+nCount;nCount++;pRoot->pRight = Tree+nCount;BuildTree(pRoot->pLeft , L , (L+R)/2);BuildTree(pRoot->pRight , (L+R)/2+1 , R);}void Insert(CNode *pRoot, int i, int v){if(pRoot->L == i && pRoot->R == i){pRoot->nSum = v;return ;}pRoot->nSum += v;if(i<=Mid(pRoot))Insert(pRoot->pLeft , i , v);elseInsert(pRoot->pRight , i , v);}void Add(CNode *pRoot, int a, int b, long long c) {if(pRoot->L == a && pRoot->R == b){pRoot->lnc += c;return;}pRoot->nSum += c*(b-a+1);if(b<=(pRoot->L + pRoot->R)/2)Add(pRoot->pLeft , a , b , c);else if(a>=(pRoot->L + pRoot->R)/2+1)Add(pRoot->pRight , a , b , c);else{Add(pRoot->pLeft , a ,(pRoot->L + pRoot->R)/2 , c);Add(pRoot->pRight , (pRoot->L + pRoot->R)/2 + 1 , b , c);}
}long long QuerynSum(CNode *pRoot, int a, int b) {if(pRoot->L == a && pRoot->R == b)return pRoot->nSum +(pRoot->R - pRoot->L + 1)*pRoot->lnc ;pRoot->nSum += (pRoot->R - pRoot->L + 1)*pRoot->lnc ;Add(pRoot->pLeft , pRoot->L , Mid(pRoot) , pRoot->lnc );Add(pRoot->pRight , Mid(pRoot) + 1 , pRoot->R , pRoot->lnc );pRoot->lnc = 0;if(b<=Mid(pRoot))return QuerynSum(pRoot->pLeft , a , b);else if(a>=Mid(pRoot)+1)return QuerynSum(pRoot->pRight , a , b);else{return QuerynSum(pRoot->pLeft , a , Mid(pRoot))+QuerynSum(pRoot->pRight , Mid(pRoot)+1 , b);}
}int main() {int n,q,a,b,c;char cmd[10];scanf("%d%d",&n,&q);int i,j,k;nCount=0;BuildTree(Tree,1,n);for(i=1; i<=n; i++){scanf("%d",&a);Insert(Tree,i,a);}for(i=0; i<q; i++){scanf("%s",cmd);if(cmd[0] == 'C'){scanf("%d%d%d",&a,&b,&c);Add(Tree,a,b,c);}else{scanf("%d%d",&a,&b);printf("%I64d\n",QuerynSum(Tree, a, b));}}return 0;
}