http://acm.hdu.edu.cn/showproblem.php?pid=4267
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
/***
hdu4267分组线段树
题目大意：对于给定区间[a,b]的数(i-a)%k==0加上c，最后单点查询。
解题思路：分组为[a,b]区间内除k余m的分为一组。这样按组来更新就好了。
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <iostream>
using namespace std;
const int maxn=50010;
int tree[maxn*4][55];
int flag[maxn*4];
int b[11][11],num[maxn];
void build(int l,int r,int root)
{
flag[root]=0;
for(int i=0;i<55;i++)
{
tree[root][i]=0;
}
if(l==r)return;
int mid=(l+r)/2;
build(l,mid,root<<1);
build(mid+1,r,root<<1|1);
}
void push_down(int root)
{
if(flag[root])
{
flag[root<<1|1]=flag[root<<1]=flag[root];
flag[root]=false;
for(int i=0;i<55;i++)
{
tree[root<<1][i]+=tree[root][i];
tree[root<<1|1][i]+=tree[root][i];
tree[root][i]=0;
}
}
}
int query(int pos,int l,int r,int root)
{
if(l==r)
{
int ret=0;
for(int i=1;i<=10;i++)
{
ret+=tree[root][b[i][pos%i]];
}
return ret;
}
push_down(root);
int m=(l+r)/2;
if(pos<=m)
return query(pos,l,m,root<<1);
else
return query(pos,m+1,r,root<<1|1);
}
void update(int mod,int L,int R,int k,int c,int l,int r,int root)
{
if(L<=l&&r<=R)
{
/// printf("*\n");
tree[root][b[k][mod]]+=c;
flag[root]=true;
return;
}
int mid=(l+r)>>1;
if(L<=mid)
update(mod,L,R,k,c,l,mid,root<<1);
if(mid<R)
update(mod,L,R,k,c,mid+1,r,root<<1|1);
}
int main()
{
int cnt=0;
for(int i=1;i<11;i++)///分组
{
for(int j=0;j<i;j++)
b[i][j]=cnt++;
}
int n,m;
while(~scanf("%d",&n))
{
build(1,n,1);
for(int i=1;i<=n;i++)
{
scanf("%d",&num[i]);
}
scanf("%d",&m);
while(m--)
{
int x;
scanf("%d",&x);
if(x==1)
{
int u,v,k,c;
scanf("%d%d%d%d",&u,&v,&k,&c);
update(u%k,u,v,k,c,1,n,1);
}
else
{
int pos;
scanf("%d",&pos);
printf("%d\n",query(pos,1,n,1)+num[pos]);
}
}
}
return 0;
}
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