A Simple Problem with Integers


Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4985    Accepted Submission(s): 1569


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​​


 


此题为区间更新，区间查询，但是在区间更新上，会有点要求，注意点就好


#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL __int64
using namespace std;
const int maxn = 50010;
struct tree
{
    int l , r, k[12];
}a[4*maxn];
int num[maxn] , N , Q;
void build(int l , int r , int k)
{
    a[k].l = l;
    a[k].r = r;
    for(int i = 0; i < 12; i++) a[k].k[i] = 0;
    if(l != r)
    {
        int mid = (l+r)/2;
        build(l , mid , 2*k);
        build(mid+1 , r , 2*k+1);
    }
}
void update(int l , int r , int k , int lk , int lc)
{
    if(l <= a[k].l && a[k].r <= r) a[k].k[lk] += lc;
    else
    {
        int mid = (a[k].l+a[k].r)/2;
        if(mid >= r) update(l , r, 2*k , lk , lc);
        else if(mid < l) update(l , r , 2*k+1 , lk , lc);
        else
        {
            update(l , mid , 2*k , lk , lc);
            int tl = mid+1+(lk-(mid+1-l)%lk)%lk;
            if(tl <= r) update(tl , r , 2*k+1 , lk , lc);
        }
    }
}
int getAns(int k , int i)
{
    int ans = 0;
    for(int j = 1; j <= 10; j++)
    {
        if((i-a[k].l)%j == 0) ans += a[k].k[j];
    }
    return ans;
}
int query(int l , int r , int k)
{
    if(l <= a[k].l && a[k].r <= r) return getAns(k , l);
    else
    {
        int mid = (a[k].l+a[k].r)/2;
        int ans = getAns(k , l);
        if(mid >= r) return ans+query(l , r , 2*k);
        else return ans+query(l , r , 2*k+1);
    }
}
int main()
{
    while(~scanf("%d" , &N))
    {
        for(int i = 1; i <= N; i++) scanf("%d" , &num[i]);
        build(1 , N , 1);
        scanf("%d" , &Q);
        int op , la, lb , lk , lc;
        while(Q--)
        {
            scanf("%d" , &op);
            if(op == 1)
            {
                scanf("%d%d%d%d" , &la , &lb , &lk , &lc);
                update(la , lb , 1 , lk , lc);
            }
            else
            {
                scanf("%d" , &la);
                printf("%d\n" , num[la]+query(la , la , 1));
            }
        }
    }
    return 0;
}