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)
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
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
1
1
1
1
3
3
1
2
3
4
1
【题意】给出n个数，再给出m个操作
1.op==1,输入四个数据a,b,k,c将区间[a,b]中的数i满足(i-a)%k == 0加上c.
2.op==2,输入一个数y,输出序列中第y个数的值。
被这题虐哭，（毕竟太弱了~~）关键就是建立多个树状数组，然而我对树状数组理解还是不行啊！
sum[x][k][x%k]代表x对k取余的值，然后每次更新树状数组的时候只需要更新update(a,.....) 与update(b+1,.....);
参考资料：http://blog.csdn.net/yeguxin/article/details/47999833
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
const int N=+;
int aa[N];
int n,m;
int sum[N][][];//开稍大一点就会MLE int lowbit(int x)
{
return x&(-x);
}
void update(int x,int k,int mod,int v)
{
while(x<=n)
{
sum[x][k][mod]+=v;
x+=lowbit(x);
}
}
int query(int x,int y)
{
int res=;
while(x)
{
for(int i=;i<=;i++)
{
res+=sum[x][i][y%i];
}
x-=lowbit(x);
}
return res;
}
int main()
{
while(~scanf("%d",&n))
{
memset(sum,,sizeof(sum));
for(int i=;i<=n;i++)
{
scanf("%d",&aa[i]);
}
scanf("%d",&m);
int op,a,b,k,c;
while(m--)
{
scanf("%d",&op);
if(op==)
{
scanf("%d",&k);
int ans=query(k,k);
printf("%d\n",ans+aa[k]);
}
else if(op==)
{
scanf("%d%d%d%d",&a,&b,&k,&c);
int kk=(b-a)/k;
update(a,k,a%k,c);
update(b+,k,a%k,-c);
}
}
}
return ;
}