标签：
|
A Simple Problem with Integers
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. 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
|
|||||||||
//11412 KB 2704 ms
#include<stdio.h>
#include<algorithm>
#include<string.h>
#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<left)update(left,right,val,2*i+1);
else 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<left){return query(left,right,i*2+1);}
else{return query(left,tree[i].mid,i*2)+query(tree[i].mid+1,right,i*2+1);}
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%I64d",&s[i]);
build(0,M,1);
char z[2];
int a,b;
ll c;
while(m--)
{
scanf("%s",z);
if(z[0]=='Q')
{
scanf("%d%d",&a,&b);
printf("%I64d\n",query(a,b,1));
}
else
{
scanf("%d%d%I64d",&a,&b,&c);
update(a,b,c,1);
}
}
}
return 0;
}
POJ 3468 A Simple Problem with Integers 线段树区间更新
标签：
原文地址：http://blog.csdn.net/crescent__moon/article/details/43531229
