POJ 3468 A Simple Problem with Integers
链接：http://poj.org/problem?id=3468
A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 77302 Accepted: 23788
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 51 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
455
9
15
Hint
The sums may exceed the range of 32-bit integers.
Source
POJ Monthly--2007.11.25, Yang Yi
大意——给你n个数字组成的序列。你将能够对它们进行两种操作，一种是对于当前给定区间里的数字都加上一个数，另一种是对于当前给定区间里的数字进行求和。
思路——表面上看起来是用线段树解决，但实际上将问题稍微处理一下就可以用树状数组来解决，而且效率更高。处理如下:如果给区间[l,r]都加上一个数x的话，我们令s(i)=加上x之前的前i项和，s’(i)=加上x之后的前i项和，那么就有当i<l时，s’(i)=s(i);当l<=i<=r时，s’(i)=s(i)+x*(i-l+1)=s(i)+x*i
-x*(l-1);当i>r时，s’(i)=s(i)+x*(r-l+1)。那么我们就可以建立两个树状数组，从而将问题简化。具体参见：csdn博客：http://blog.csdn.net/u013068502/article/details/47252335。
-x*(l-1);当i>r时，s’(i)=s(i)+x*(r-l+1)。那么我们就可以建立两个树状数组，从而将问题简化。具体参见：csdn博客：http://blog.csdn.net/u013068502/article/details/47252335。
复杂度分析——时间复杂度:O(log(n!)+q*log(ab)),空间复杂度:O(n)
附上AC代码：
#include <iostream>
#include <cstdio>
#include <string>
#include <cmath>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <set>
#include <map>
using namespace std;
typedef unsigned int UI;
typedef long long LL;
typedef unsigned long long ULL;
typedef long double LD;
const double pi = acos(-1.0);
const double e = exp(1.0);
const double eps = 1e-8;
const int maxn = 100005;
LL bit1[maxn], bit2[maxn]; // 代表两个树状数组，数组1用来存储初始值，
// 数组2用来存储变化的值
int n, query; // 初始数组大小，问题的个数
char op[5]; // 选择哪种操作
int lowbit(int x);
void update(LL * bit, int x, int add);
LL sum(LL * bit, int x);
int main()
{
ios::sync_with_stdio(false);
int a, b, c, x;
while (~scanf("%d%d", &n, &query))
{
memset(bit1, 0, sizeof(bit1));
memset(bit2, 0, sizeof(bit2));
// 上面清空数组，避免上次结果干扰
for (int i=1; i<=n; i++)
{
scanf("%d", &x);
update(bit1, i, x); // 将初始值存入数组1
}
while (query--)
{
scanf("%s%d%d", op, &a, &b);
if (op[0] == 'C')
{
scanf("%d", &c);
update(bit1, a, (-c)*(a-1));
update(bit2, a, c);
// 上面更新改变状态
update(bit1, b+1, c*b);
update(bit2, b+1, (-c));
// 上面去掉重复状态
}
else
{
LL ans = 0;
ans += sum(bit1, b)+sum(bit2, b)*b;
ans -= sum(bit1, a-1)+sum(bit2, a-1)*(a-1);
printf("%lld\n", ans);
}
}
}
return 0;
}
int lowbit(int x) // 求2^k，k表示x为二进制时末尾的0的个数
{
return (x&(-x));
}
void update(LL * bit, int x, int add)
{ // 更新节点信息(加上数add)
while (x!=0 && x<=n)
{
bit[x] += add;
x += lowbit(x);
}
}
LL sum(LL * bit, int x)
{ // 区间求和(求1~x之间数组的和)
LL res = 0;
while (x > 0)
{
res += bit[x];
x -= lowbit(x);
}
return res;
}