POJ3468 A Simple Problem with Integers
/*线段树区间更新
2000MS 4484K
*/
#include <stdio.h>
#include <string.h>
const int MAX=1e5+5;
long long tree[MAX*4];
long long lazy[MAX*4];//延时数组
void read(int &x)//读入优化
{
long long f=1;x=0;char s=getchar();
while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
while(s>='0'&&s<='9'){x=x*10+s-'0';s=getchar();}
x*=f;
}
void pushup (int root)//向上更新
{
tree[root]=tree[root*2]+tree[root*2+1];
}
void pushdown (int root,int left,int right,int mid)//向下更新
{
tree[root*2]+=lazy[root]*(mid-left+1);
tree[root*2+1]+=lazy[root]*(right-mid);
lazy[root*2]+=lazy[root];
lazy[root*2+1]+=lazy[root];
lazy[root]=0;
}
void build (int root,int left,int right)//建树
{
lazy[root]=0;
if (left==right)
{
int x;
read(x);
tree[root]=x;
return ;
}
int mid=(left+right)/2;
build(root*2,left,mid);
build(root*2+1,mid+1,right);
pushup(root);
}
void update (int root,int left,int right,int L,int R,int val)//区间更新
{
if (L<=left&&R>=right)
{
tree[root]+=(right-left+1)*val;
lazy[root]+=val;
return ;
}
int mid=(right+left)/2;
if (lazy[root]!=0) pushdown(root,left,right,mid);
if (L<=mid) update(root*2,left,mid,L,R,val);
if (R>mid) update(root*2+1,mid+1,right,L,R,val);
pushup(root);
}
long long query (int root,int left,int right,int L,int R)//区间查询
{
if (L<=left&&R>=right)
return tree[root];
int mid=(right+left)/2;
long long sum=0;
if (lazy[root]!=0) pushdown(root,left,right,mid);
if (L<=mid) sum+=query(root*2,left,mid,L,R);
if (R>mid) sum+=query(root*2+1,mid+1,right,L,R);
return sum;
}
int main ()
{
int n,m;
read(n);read(m);
build(1,1,n);
while (m--)
{
int x,y,z;
char c;
c=getchar();
if (c=='Q')
{
read(x);read(y);
printf ("%I64d\n",query(1,1,n,x,y));//long long 64位
}
else
{
read(x);read(y);read(z);
update(1,1,n,x,y,z);
}
}
return 0;
}