线段树 POJ 3468 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.
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 abc" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q ab" 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
#include
#include
#include
using namespace std;
long long a[100217*4];
long long col[100217*4];
int temp=1;
void build(int rt, int begin, int end)
{
int m=(begin+end)/2;
if(begin==end)
{
scanf("%I64d",&a[rt]);
return ;
}
build( rt << 1, begin, m);
build( 1 + (rt << 1), m + 1,end);
a[rt] = a[rt << 1] + a[1 + (rt << 1)];
}
void update(int rt, int begin, int end, int x, int y, int v)
{
int m=(begin+end)/2;
if(x>end||y=x)
{
a[rt]+=(end-begin+1)*v;
col[rt]+=v;
//printf("%d %d %d %d %d\n",col[rt], begin, end, rt, a[rt]);
return ;
}
if(col[rt]!=0)
{
a[rt<<1] += (m-begin+1)*col[rt];
a[(rt<<1)+1] += (end-(m+1)+1)*col[rt];
col[rt<<1] += col[rt] ;
col[(rt<<1)+1] += col[rt] ;
col[rt] = 0;
}
update( rt << 1, begin, m, x, y, v);
update( 1 + (rt << 1), m + 1,end, x, y, v);
a[rt] = a[rt << 1] + a[1 + (rt << 1)];
}
long long query(int rt, int begin, int end, int x, int y)
{
int m=(begin+end)/2;
if(x>end||y=x)
{
//printf("%d-\n",rt);
//printf("%d\n",a[rt]);
long long t=a[rt];
return t;
}
if(col[rt]!=0)
{
a[rt<<1 ]+= (m-begin+1)*col[rt];
a[(rt<<1)+1] += (end-(m+1)+1)*col[rt];
col[rt<<1] += col[rt] ;
col[(rt<<1)+1] += col[rt] ;
col[rt] = 0;
}
long long p1 = 0,p2 = 0;
p1 = query( rt << 1, begin, m, x, y);
p2 = query( 1 + (rt << 1), m + 1,end, x, y);
//printf("%d %d\n",p1,p2);
return p1+p2;
//a[rt]=a[rt << 1] + a[1 + (rt << 1)];
}
int main()
{
int T,n,m;
while(scanf("%d%d",&n,&m)!=-1)
{
build(1,1,n);
// puts("1");
memset(col,0,sizeof(col));
// getchar();
// printf("%d\n",a[1]);
while(m--)
{
//getchar();
int x,y,v;
char ch[10];
scanf("%s",&ch);
//printf("%c\n",ch);
if(ch[0]=='Q')
{
scanf("%d%d", &x, &y);
printf("%lld\n", query(1,1,n,x,y));
// puts("y");
// for(int i=1;i