POJ 3468
A Simple Problem with Integers
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total Submissions: 46936 | Accepted: 13755 | |
| 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
线段树模板题，区间更新和区间求和。
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
using namespace std;
//#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , m , pos << 1
#define rson m + 1 , r , pos << 1|1
#define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)
#define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)
#define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)
#define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)
const int MAXN = 100000 + 100;
const int maxw = 10000000 + 20;
const int MAXNNODE = 1000000 +10;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
#define eps 1e-8
#define mod 1000000007
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pi;
int n , m , x , y ;
char ch[5];
LL sum[MAXN << 2] , lazy[MAXN << 2];
void push_up(int pos)
{
sum[pos] = sum [pos << 1] + sum[pos << 1|1];
}
void push_down(int l , int r , int pos)
{
if(lazy[pos])
{
lazy[pos << 1] += lazy[pos];
lazy[pos << 1|1] += lazy[pos];
sum[pos << 1] += lazy[pos] * (r - l + 1 - (( r - l + 1) >> 1));
sum[pos << 1|1] += lazy[pos] * ((r - l + 1) >> 1);
lazy[pos] = 0;
}
}
void build(int l , int r , int pos)
{
if(l == r)
{
scanf("%lld" , &sum[pos]);
}
else
{
int m = (l + r) >> 1;
build(lson);
build(rson);
push_up(pos);
}
}
void update(int L , int R , int l , int r , int pos , long long number)
{
if(L <= l&&r <= R)
{
lazy[pos] += number;
sum[pos] += number * (r - l + 1);
return;
}
push_down(l , r ,pos);
int m = (l + r) >> 1;
if(L <= m)update(L , R , lson , number);
if(R > m)update(L , R , rson , number);
push_up(pos);
}
LL querysum(int L , int R , int l , int r , int pos)
{
if(L <= l&&R >= r)
{
return sum[pos];
}
push_down(l , r , pos);
LL ans = 0;
int m = (l + r) >> 1;
if(L <= m)ans += querysum(L , R , lson);
if(R > m)ans += querysum(L , R ,rson);
return ans;
}
int main()
{
//ios::sync_with_stdio(false);
#ifdef Online_Judge
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif // Online_Judge
while(scanf("%d%d" , &n ,&m) == 2)
{
build(1 , n , 1);
clr(lazy , 0);
while(m--)
{
scanf("%s" , ch);
if(ch[0] == 'Q')
{
scanf("%d%d" , &x , &y);
printf("%lld\n" , querysum(x , y , 1 , n , 1));
}
else
{
LL num;
scanf("%d%d%lld" , &x , &y , &num);
update(x , y , 1 , n , 1 , num);
}
}
}
return 0;
}