A Simple Problem with Integers
区间更新，区间求和
http://poj.org/problem?id=3468
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 87351 | Accepted: 27114 | |
| Case Time Limit: 2000MS | ||
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
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <string>
#include <iostream>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
const int SIZE=1e5+10;
using namespace std;
struct sge{
LL sum;
LL lazy;
}a[SIZE<<2];
void pushup(int rt){
a[rt].sum=a[rt<<1].sum+a[rt<<1|1].sum;
}
void pushdown(int l,int r,int rt){
if(a[rt].lazy){
int m=(r-l+1);
a[rt<<1].sum+=(m-m/2)*a[rt].lazy;
a[rt<<1|1].sum+=(m/2)*a[rt].lazy;
a[rt<<1].lazy+=a[rt].lazy;
a[rt<<1|1].lazy+=a[rt].lazy;
a[rt].lazy=0;
}
}
void build(int l,int r,int rt){
a[rt].lazy=0;
if(l==r){
scanf("%lld",&a[rt].sum);
return;
}
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt){
if(L<=l&&r<=R){
a[rt].sum+=(r-l+1)*c;
a[rt].lazy+=c;
return;
}
pushdown(l,r,rt);
int m=(l+r)>>1;
if(L<=m)update(L,R,c,lson);
if(R>m)update(L,R,c,rson);
pushup(rt);
}
LL query(int L,int R,int l,int r,int rt){
if(L<=l&&r<=R){
return a[rt].sum;
}
pushdown(l,r,rt);
int m=(l+r)>>1;
LL sum=0;
if(L<=m)sum+=query(L,R,lson);
if(R>m)sum+=query(L,R,rson);
return sum;
}
int main()
{
int n,q,x,y,c;
char s[10];
while(scanf("%d%d",&n,&q)!=EOF){
build(1,n,1);
for(int i=0;i<q;i++){
scanf("%s",s);
if(s[0]=='Q'){
scanf("%d%d",&x,&y);
printf("%lld\n",query(x,y,1,n,1));
}
else {
scanf("%d%d%d",&x,&y,&c);
update(x,y,c,1,n,1);
}
}
}
return 0;
}