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
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
 

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 = ;
int tree[MAXN][][];
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 = ;
while(x > ){
for(int i = ; i <= ; i++){
ans += tree[x][i][a % i];
}
x -= lowbit(x);
}
return ans;
}
int main(){
int N, M;
while(~scanf("%d", &N)){
memset(tree, , sizeof(tree));
for(int i = ; 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 == ){
scanf("%d%d%d", &b, &k, &c);
add(a, k, a % k, c);
add(b + , k, a % k, -c);
}
else{
printf("%d\n", num[a] + sum(a, a));
}
}
}
return ;
}

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