poj 3468 A Simple Problem with Integers（线段树区间更新求和）

题目地址

题目大意：给出n长度的数组

"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.

解题思路：裸线段树区间更新并求和，注意用long long

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>

using namespace std;

typedef __int64 LL;
const int maxn = 100000+100;

struct Node
{
int l,r;
LL add,sum;
}tree[4*maxn];
int a[maxn];

void build(int id,int l,int r)
{
tree[id].l = l;
tree[id].r = r;
tree[id].add = 0;
if(l == r)
{
tree[id].sum = a[l];
return;
}
int mid = (tree[id].l+tree[id].r)/2;
build(2*id,l,mid);
build(2*id+1,mid+1,r);
tree[id].sum = tree[id*2].sum+tree[id*2+1].sum;
}

void update(int id,int l,int r,LL c)
{
if(tree[id].l == l && tree[id].r == r)///区间对应
{
tree[id].add += c;
return;
}
tree[id].sum += LL(r-l+1)*c;
int mid = (tree[id].l+tree[id].r)/2;
if(r <= mid)
update(id*2,l,r,c);
else if(l >= mid+1)
update(id*2+1,l,r,c);
else
{
update(id*2,l,mid,c);
update(id*2+1,mid+1,r,c);
}
}

LL query(int id,int l,int r)///查询l到r区间的和
{
if(tree[id].l == l && tree[id].r == r)
return tree[id].sum+tree[id].add*(r-l+1);
if(tree[id].add)
{
tree[id*2].add += tree[id].add;
tree[id*2+1].add += tree[id].add;
tree[id].sum += LL(tree[id].r-tree[id].l+1)*tree[id].add;
tree[id].add = 0;
}
int mid = (tree[id].l+tree[id].r)/2;
if(r <= mid)
return query(id*2,l,r);
else if(l >= mid+1)
return query(id*2+1,l,r);
else
return query(id*2,l,mid)+query(id*2+1,mid+1,r);
}

int main()
{
int n,m;
while(scanf("%d%d",&n,&m) != EOF)
{
for(int i = 1; i <= n; i++)
scanf("%d\n",&a[i]);
build(1,1,n);
char ch[10];
int a,b,c;
for(int i = 1; i <= m; i++)
{
scanf("%s",ch);
if(ch[0] == 'Q')
{
scanf("%d%d", &a,&b);
printf("%I64d\n",query(1,a,b));
}
else
{
scanf("%d%d%d",&a,&b,&c);
update(1,a,b,c);
}
}
}
return 0;
}