HDU 4267

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 323 Accepted Submission(s): 133

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

//原来我用插点问线做的，输出sum(temp) - sum(temp-1),一直tle 
#include <stdio.h>
#include <iostream>
#include<string.h>
using namespace std;

const int MAXD = 50010;
int N, M, d[100][MAXD], a[MAXD];

void insert(int k, int x, int v)
{
     for(; x <= N; x += x & -x) 
          d[k][x] += v;
}

void init()
{
    int i, j;
     for(i = 1; i <= N; i ++)
          cin>>a[i];
     for(i = 0; i < 100; i ++) 
          memset(d[i], 0, sizeof(d[i][0]) * (N + 1));
}

int query(int id)
{
    int i, k, x, ans = 0;
    for(i = 1; i <= 10; i ++)
    {
        k = (i - 1) * 10 + id % i;
        for(x = id; x > 0; x -= x & -x) 
          ans += d[k][x];
    }
    return a[id] + ans;
}

void solve()
{
    int i, flag,temp;
    int a, b, k, c;
    cin>>M;
    while(M--)
    {
        cin>>flag;
        if(flag == 1)
        {
            cin>>a>>b>>k>>c;
            b -= (b - a) % k;
            insert(10 * (k - 1) + a % k, a, c);
            if(b < N) insert(10 * (k - 1) + b % k, b + 1, -c);
        }
        else if(flag == 2)
        {
            cin>>temp;
            cout<<query(temp)<<endl;
        }
    }
}

int main()
{
    while(cin>>N)
    {
        init();
        solve();
    }
    return 0;
}