集训队专题（10）1002 A Simple Problem with Integers

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本文介绍了一种涉及区间更新和查询的算法问题，通过构建特殊的数据结构实现高效处理。包括了对特定区间内的元素进行批量修改及快速获取指定位置元素值的方法。

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

Sample Output

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;
}