POJ3468 —— 线段树 || 树状数组
|
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. 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
|
|||||||||
题意是给你一个数列，有两种操作：1。C a b c 把区间a到b的每个元素都加上c ；2。Q a b 输出区间a到b的和。
思路：1.线段树，要是用线段树做的话这个就是裸的区间修改+区间查询 ， 维护区间和即可。
2.树状数组，这种属于一类比较巧妙的树状数组。
我们开两个树状数组，bit0 ， bit1；我们设a数组从i到j的和 = bit0的前i项和 * i + bit1的前i项和。
那么我们把区间（l ， r）加上add可以等效为：
1.在bit0的l上加- x * （l - 1）；
2.在bit1的l上加x；
3.在bit0的r + 1上加x * r；
4.在bit1的r + 1上加- x；
这样就可以用树状数组了。PS：注意要用long long
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <map>
#include <string>
#include <stack>
#include <cctype>
#include <vector>
#include <queue>
#include <set>
#include <utility>
#include <cassert>
using namespace std;
///#define Online_Judge
#define outstars cout << "***********************" << endl;
#define clr(a,b) memset(a,b,sizeof(a))
#define lson l , mid , rt << 1
#define rson mid + 1 , r , rt << 1 | 1
#define mk make_pair
#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 + 5;
const int MAXS = 10000 + 50;
const int sigma_size = 26;
const long long LLMAX = 0x7fffffffffffffffLL;
const long long LLMIN = 0x8000000000000000LL;
const int INF = 0x7fffffff;
const int IMIN = 0x80000000;
const int inf = 1 << 30;
#define eps 1e-8
const long long MOD = 1000000000 + 7;
const int mod = 100000;
typedef long long LL;
const double PI = acos(-1.0);
typedef double D;
typedef pair<int , int> pii;
#define Bug(s) cout << "s = " << s << endl;
///#pragma comment(linker, "/STACK:102400000,102400000")
int n , q , a[MAXN] ;
LL bit0[MAXN] , bit1[MAXN];
LL lowbit(int x){return x & (-x);}
void update(LL *c , int x , int add)
{
while(x <= MAXN)
{
c[x] += add;
x += lowbit(x);
}
}
LL getsum(LL *c , int x)
{
LL sum = 0 ;
while(x > 0)///不能有等于零的情况
{
sum += c[x];
x -= lowbit(x);
}
return sum;
}
int main()
{
while(~scanf("%d%d" , &n , &q))
{
clr(bit0 ,0) , clr(bit1 , 0);
for(int i = 1 ; i <= n ; i++)
{
scanf("%d" , &a[i]);
// outstars
update(bit0 , i , a[i]);
}
char cmd[5];
// outstars
for(int i = 0 ; i < q ; i++)
{
// Bug(q)
// getchar();
int a , b , c;
scanf("%s%d%d" , cmd , &a , &b);
// Bug(cmd)
// outstars
if(cmd[0] == 'C')
{
// int a , b , c;
scanf("%d" , &c);
update(bit0 , a , (-c) * (a - 1));
update(bit1 , a , c);
update(bit0 , b + 1 , b * c);
update(bit1 , b + 1 , -c);
}
else
{
// int a , b;
// scanf("%d%d" , &a ,&b);
// outstars
LL ans = 0;
// Bug(getsum(bit0 , b));
ans += getsum(bit0 , b) + getsum(bit1 , b) * b;
// cout << "999" << ' ' << endl;
ans -= getsum(bit0 , a - 1) + getsum(bit1 , a - 1) * (a - 1);
printf("%lld\n" , ans);
}
// outstars
}
}
return 0;
}