A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 92127 | Accepted: 28671 | |
| Case Time Limit: 2000MS | ||
描述
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.
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、
前几次提交的时候没有吧node[u].add清零，一直WA
第一道线段树的题，当晚没有调出来，第二天才想出来，不错
#include<cstdio>
#include<iostream>
#define LL long long
#define R(u) (u<<1|1)
#define L(u) (u<<1)
using namespace std;
const int maxx=100005;
LL a[maxx];
int n,m;
struct Node{
int r,l;
LL add,sum;
}node[maxx<<2];
void Pushup(int u)
{
node[u].sum=node[L(u)].sum+node[R(u)].sum;
return;
}
void Pushdown(int u)
{
node[L(u)].add+=node[u].add;
node[R(u)].add+=node[u].add; node[L(u)].sum+=(node[L(u)].r-node[L(u)].l+1)*node[u].add; node[R(u)].sum+=(node[R(u)].r-node[R(u)].l+1)*node[u].add;
node[u].add=0; //一定记得清零
}
void Build(int u,int left,int right)
{
node[u].l=left,node[u].r=right;
node[u].add=0;
if(left==right)
{
node[u].sum=a[left];
return;
}
int mid=(left+right)>>1;
Build(L(u),left,mid);
Build(R(u),mid+1,right);
Pushup(u);
}
void update(int u,int left,int right,LL val)
{
if(left==node[u].l&&node[u].r==right)
{
node[u].add+=val;
node[u].sum+=(node[u].r-node[u].l+1)*val;
return;
}
node[u].sum+=(right-left+1)*val;// 当更新区间小于这段
int mid=(node[u].r+node[u].l)>>1;
if(mid>=right) update(L(u),left,right,val);//在左边
else if(mid<left) update(R(u),left,right,val);
else {
update(L(u),left,mid,val);
update(R(u),mid+1,right,val);
}
//Pushup(u);前面已经直接算出sum后面不用再Pushup了
}
LL Qurey(int u,int left,int right)
{
if(left==node[u].l&&node[u].r==right)
return node[u].sum;
if(node[u].add)Pushdown(u);
int mid=(node[u].r+node[u].l)>>1;
if(mid>=right) Qurey(L(u),left,right);
else if(mid<left) Qurey(R(u),left,right);
else return (Qurey(L(u),left,mid)+Qurey(R(u),mid+1,right));
//Pushup(u);
}
int main()
{
cin>>n>>m;
LL c;
for(int i=1;i<=n;i++)
scanf("%I64d",a+i);
Build(1,1,n);
while(m--)
{char x;int ai,an;
scanf("%c %d %d",&x,&ai,&an);
cin>>x>>ai>>an;
if(x=='C')
{
scanf("%I64d",&c);
update(1,ai,an,c);
}
else
printf("%I64d\n",Qurey(1,ai,an));
}
return 0;
}