A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 149972 | Accepted: 46526 |
题目链接：http://poj.org/problem?id=3468
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
题解：
线段树模板题，注意一下lazy标记的下传操作，标记也是long long 型的。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;
const int N = 1e5+;
int n,m;
ll a[N];
ll ans;
struct Tree{
int l,r;
ll f,w;
}tre[(N<<)+];
void build(int o,int l,int r){
tre[o].l=l;tre[o].r=r;tre[o].f=;
if(l==r){
tre[o].w=a[l];
return ;
}
int mid=l+r>>;
build(o<<,l,mid);
build(o<<|,mid+,r);
tre[o].w=tre[o<<].w+tre[o<<|].w;
}
void down(int o){
tre[o<<].f+=tre[o].f;
tre[o<<|].f+=tre[o].f;
tre[o<<].w+=tre[o].f*(tre[o<<].r-tre[o<<].l+);
tre[o<<|].w+=tre[o].f*(tre[o<<|].r-tre[o<<|].l+);
tre[o].f=;
}
void update(int o,int l,int r,int val){
int L=tre[o].l,R=tre[o].r;
if(L>=l && R<=r){
tre[o].w+=(ll)val*(R-L+);
tre[o].f+=val;
return ;
}
down(o);
int mid=L+R>>;
if(l<=mid) update(o<<,l,r,val);
if(r>mid) update(o<<|,l,r,val);
tre[o].w=tre[o<<].w+tre[o<<|].w;
}
void query(int o,int l,int r){
int L=tre[o].l,R=tre[o].r;
if(L>=l && R<=r){
ans+=tre[o].w;
return ;
}
down(o);
int mid=L+R>>;
if(l<=mid) query(o<<,l,r);
if(r>mid) query(o<<|,l,r);
}
int main(){
scanf("%d%d",&n,&m);
for(int i=;i<=n;i++) scanf("%I64d",&a[i]);
build(,,n);
char s[];
for(int i=;i<=m;i++){
scanf("%s",s);
if(s[]=='Q'){
int l,r;ans=;
scanf("%d%d",&l,&r);
query(,l,r);
printf("%I64d\n",ans);
}else{
int a,b,c;
scanf("%d%d%d",&a,&b,&c);
update(,a,b,c);
}
}
return ;
}