POJ 3468A Simple Problem with Integers(线段树区间更新)
A Simple Problem with Integers
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 112228 | Accepted: 34905 | |
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 题意就是如题,对线段树的区间进行更新,可增可减可修改,并查询区间和,此题就是增,具体实现看代码吧.
AC代码:
Source Code
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
#define maxsize 100005
#define LL long long
LL num[maxsize];
struct node
{
LL l,r;
LL maxn,add;
int mid()
{
return (l+r)/;
}
} tree[maxsize<<];
void Build_Tree(LL root,LL l,LL r)
{
tree[root].l=l;
tree[root].r=r;
tree[root].add=;
if(l==r)
{
tree[root].maxn=num[l];
return ;
}
Build_Tree(root*,l,tree[root].mid());
Build_Tree(root*+,tree[root].mid()+,r);
tree[root].maxn=tree[root*].maxn+tree[root*+].maxn;
}
void Update(LL root,LL l,LL r,LL czp)
{
if(tree[root].l==l&&tree[root].r==r)
{
tree[root].add+=czp;
return ;
}
tree[root].maxn+=((r-l+)*czp);
if(tree[root].mid()>=r) Update(root<<,l,r,czp);
else if(tree[root].mid()<l) Update(root<<|,l,r,czp);
else
{
Update(root<<,l,tree[root].mid(),czp);
Update(root<<|,tree[root].mid()+,r,czp);
}
}
long long Query(LL root,LL l,LL r)
{
if(tree[root].l==l&&tree[root].r==r)
{
return tree[root].maxn+(r-l+)*tree[root].add;
}
tree[root].maxn+=(tree[root].r-tree[root].l+)*tree[root].add;
Update(root<<,tree[root].l,tree[root].mid(),tree[root].add);
Update(root<<|,tree[root].mid()+,tree[root].r,tree[root].add);
tree[root].add=;
if(tree[root].mid()>=r) return Query(root*,l,r);
else if(tree[root].mid()<l) return Query(root*+,l,r);
else
{
LL Lans=Query(root*,l,tree[root].mid());
LL Rans=Query(root*+,tree[root].mid()+,r);
return Lans+Rans;
}
}
int main()
{
/*ios::sync_with_stdio(false);*/
char str[];
LL m,n,a,b,c;
while(scanf("%lld%lld",&m,&n)!=EOF)
{
for(LL i=; i<=m; i++) scanf("%I64d",&num[i]);
Build_Tree(,,m);
for(LL i=; i<=n; i++)
{
scanf("%s%I64d%I64d",str,&a,&b);
if(str[]=='Q') printf("%I64d\n",Query(,a,b));
else if(str[]=='C')
{
scanf("%I64d",&c);
Update(,a,b,c);
}
}
}
return ;
}
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