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 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
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
线段树功能:update:成段增减 query:区间求和
#include<cstdio>
#include<algorithm> #define clr(x,y) memset(x,y,sizeof(x))
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1 const int maxn=1e5+;
using namespace std; LL sum[maxn<<],Lazy[maxn<<]; void PushUp(int rt)
{
sum[rt]=sum[rt<<]+sum[rt<<|];
} void PushDown(int rt,int m)
{
if(Lazy[rt]) {
Lazy[rt<<]+=Lazy[rt];
Lazy[rt<<|]+=Lazy[rt];
sum[rt<<]+=(m-(m>>))*Lazy[rt];
sum[rt<<|]+=(m>>)*Lazy[rt];
Lazy[rt]=;
}
} void build(int l,int r,int rt)
{
int m;
Lazy[rt]=;
if(l==r) {
scanf("%lld",&sum[rt]);
return;
} m=(l+r)>>;
build(lson);
build(rson);
PushUp(rt);
} void Updata(int L,int R,int c,int l,int r,int rt)
{
int m;
if(L<=l && r<=R) {
Lazy[rt]+=c;
sum[rt]+=(LL)c*(r-l+);
return;
} PushDown(rt,r-l+);
m=(l+r)>>;
if(L<=m) Updata(L,R,c,lson);
if(R>m) Updata(L,R,c,rson);
PushUp(rt); } LL query(int L,int R,int l,int r,int rt)
{
int m;
LL ret=;
if(L<=l && r<=R) {
return sum[rt];
} PushDown(rt,r-l+);
m=(l+r)>>;
if(L<=m) ret+=query(L,R,lson);
if(R>m) ret+=query(L,R,rson); return ret;
} int main()
{
int Q,n,a,b,c;
char st[]; scanf("%d%d",&n,&Q);
build(,n,); while(Q--) {
scanf("%s",st);
if(st[]=='C') {
scanf("%d%d%d",&a,&b,&c);
Updata(a,b,c,,n,);
} else {
scanf("%d%d",&a,&b);
printf("%lld\n",query(a,b,,n,));
} } return ;
}