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A Simple Problem with Integers
| Time Limit: 5000MS |
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Memory Limit: 131072K |
| Total Submissions: 67718 |
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Accepted: 20897 |
| 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 The sums may exceed the range of 32-bit integers. Source POJ Monthly--2007.11.25, Yang Yi |
给你n个数，Q（a,b)代表查询区间[a,b]数字之和
C（a,b,c)代表在区间[a,b]增加值c
在点更新的基础上加上一个mark标记。
//11412 KB 2704 ms
#include
#include
#include
#define ll __int64 #define M 100007 using namespace std; struct node { ll l,r,mid,val,mark; }tree[M<<2]; ll s[M]; void build(ll left,ll right,ll i)//建树 { tree[i].l=left;tree[i].r=right; tree[i].mid=(left+right)>>1;tree[i].mark=0; if(left==right){tree[i].val=s[left]; return;} build(left,tree[i].mid,i*2); build(tree[i].mid+1,right,i*2+1); tree[i].val=tree[i*2].val+tree[i*2+1].val; } void update(int left,int right,ll val,int i)//区间更新 { if(tree[i].l==left&&tree[i].r==right){tree[i].mark+=val;return;} tree[i].val+=val*(right-left+1); if(tree[i].mid
=right)update(left,right,val,2*i); else { update(left,tree[i].mid,val,2*i); update(tree[i].mid+1,right,val,2*i+1); } } ll query(int left,int right,int i)//区间查询 { if(tree[i].l==left&&tree[i].r==right) return tree[i].val+tree[i].mark*(right-left+1); if(tree[i].mark!=0) { tree[i*2].mark+=tree[i].mark;tree[i*2+1].mark+=tree[i].mark; tree[i].val+=(tree[i].r-tree[i].l+1)*tree[i].mark;tree[i].mark=0; } if(tree[i].mid>=right){return query(left,right,i*2);} else if(tree[i].mid