A Simple Problem with Integers（100棵树状数组）

A Simple Problem with Integers

Time Limit: 5000/1500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5034 Accepted Submission(s): 1589

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

题解：树状数组，每次更新(i - a)%k = 0 的位置的值，每次询问a位置的值；所以i%k = a%k;

开100棵树状数组，tree[x][k][mod];

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
const int MAXN = 50100;
int tree[MAXN][11][11];
int num[MAXN];
int lowbit(int x){return x & (-x);}
void add(int x, int k, int mod, int v){
    while(x < MAXN){
        tree[x][k][mod] += v;
        x += lowbit(x);
    }
}
int sum(int x, int a){
    int ans = 0;
    while(x > 0){
        for(int i = 1; i <= 10; i++){
            ans += tree[x][i][a % i];
        }
        x -= lowbit(x);
    }
    return ans;
}
int main(){
    int N, M;
    while(~scanf("%d", &N)){
        memset(tree, 0, sizeof(tree));
        for(int i = 1; i <= N; i++){
            scanf("%d", &num[i]);
        }
        scanf("%d", &M);
        int q, a, b, k, c;
        while(M--){
            scanf("%d%d", &q, &a);
            if(q == 1){
                scanf("%d%d%d", &b, &k, &c);
                add(a, k, a % k, c);
                add(b + 1, k, a % k, -c);
            }
            else{
                printf("%d
", num[a] + sum(a, a));
            }
        }
    }
    return 0;
}