HDU4267 A Simple Problem with Integers（树状数组 离散化）

A Simple Problem with Integers

题目链接

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5991 Accepted Submission(s): 1919

Problem Description

Let A1, A2, … , AN be N elements. You need to deal with two kinds of operations. One type of operation is to add a given number to a few numbers in a given interval. The other is to query the value of some element.

Input

There are a lot of test cases.

The first line contains an integer N. (1 <= N <= 50000)

The second line contains N numbers which are the initial values of A1, A2, … , AN. (-10,000,000 <= the initial value of Ai <= 10,000,000)

The third line contains an integer Q. (1 <= Q <= 50000)

Each of the following Q lines represents an operation.

“1 a b k c” means adding c to each of Ai which satisfies a <= i <= b and (i - a) % k == 0. (1 <= a <= b <= N, 1 <= k <= 10, -1,000 <= c <= 1,000)

“2 a” means querying the value of Aa. (1 <= a <= N)

Output

For each test case, output several lines to answer all query operations.

Sample Input

4

1 1 1 1

14

2 1

2 2

2 3

2 4

1 2 3 1 2

2 1

2 2

2 3

2 4

1 1 4 2 1

2 1

2 2

2 3

2 4

Sample Output

1

1

1

1

1

3

3

1

2

3

4

1

Source

2012 ACM/ICPC Asia Regional Changchun Online

题意：给一些数，有两种操作，一种是在[a,b] 区间内，对（i - a）% k == 0 的加value,另一种操作是询问某个位置的值。

思路：区间操作，离散化思想去写。

不来想着用线段树写的，可是会内存超限，看大佬用树状数组写的，开55棵树状数组，用cnt[x][k][mod] 表示x 对k 取余为mod，这样在询问的时候只要循环到10就可以 了。而且(i - a) % k == 0即是 i % k == a % k == mod,这样更新的时候就可以更新(1,b) 和（1，a）了。

随后用线段树写一下再补上。

/*
区间修改树状数组 和离散化

*/
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namesp
4000
ace std;
const int mm=50010;
int cnt[mm][11][11];
int num[mm];
int lowbit(int x)
{
return x&(-x);
}
void update(int x,int k,int mod,int add)//对a%K 进行修改存储
{
while(x>0)
{
cnt[x][k][mod]+=add;
x-=lowbit(x);
}
}
int sum(int x,int a)
{
int s=0;
while(x<mm)
{
for(int i=1;i<=10;i++)//k小于等于10
s+=cnt[x][i][a%i];
x+=lowbit(x);
}
return s;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&num[i]);
memset(cnt,0,sizeof(cnt));
int m;
scanf("%d",&m);
while(m--)
{
int o,a,b,k,c;
scanf("%d",&o);
if(o==1)
{
scanf("%d%d%d%d",&a,&b,&k,&c);
update(b,k,a%k,c);
update(a-1,k,a%k,-1*c);//由于b 算的是 b到0的  需要修改的树状数组树 a到 b  所以要对a前面的树状数组进行修改，以免影响后面的操作
}
else
{
scanf("%d",&a);
int ans=sum(a,a);
printf("%d\n",ans+num[a]);
}
}
}
return 0;
}