1.题目链接。题目的操作其实很像去区间修改,区间查询的操作了,但是这个操作的区间不再是连续的了,而是离散的。但是这里还是可以用树状数组维护的,因为k很小,所以k和i%k其实只有C(10,2)种情况,我们对每一种情况建一颗树状数组,然后查询的时候查对应的树状数组即可。
#include<bits/stdc++.h>
using namespace std;
#define lowbit(x) (x)&(-x)
const int MAXN = 50000 + 5;
int num[MAXN];
int c[MAXN][11][11];
int n;
void add(int x, int k, int mod, int val)
{
for (int i = x; i <= n; i += lowbit(i))
c[i][k][mod] += val;
}
int sum(int x, int a)
{
int ret = 0;
for (int i = x; i > 0; i -= lowbit(i))
for (int j = 1; j <= 10; ++j)
ret += c[i][j][a%j];
return ret;
}
int main()
{
while (~scanf("%d", &n)) {
for (int i = 1; i <= n; ++i)
scanf("%d", &num[i]);
memset(c, 0, sizeof(c));
int q, a, b, k, c, op;
scanf("%d", &q);
while (q--)
{
scanf("%d", &op);
if (op == 1)
{
scanf("%d%d%d%d", &a, &b, &k, &c);
add(a, k, a%k, c);
add(b + 1, k, a%k, -c);
}
else {
scanf("%d", &a);
printf("%d\n", num[a] + sum(a, a));
}
}
}
return 0;
}