CF GYM 100741 A Queries

A. Queries
time limit per test0.25 seconds
memory limit per test64 megabytes
inputstandard input
outputstandard output
Mathematicians are interesting (sometimes, I would say, even crazy) people. For example, my friend, a mathematician, thinks that it is very fun to play with a sequence of integer numbers. He writes the sequence in a row. If he wants he increases one number of the sequence, sometimes it is more interesting to decrease it (do you know why?..) And he likes to add the numbers in the interval [l;r]. But showing that he is really cool he adds only numbers which are equal some mod (modulo m).

Guess what he asked me, when he knew that I am a programmer? Yep, indeed, he asked me to write a program which could process these queries (n is the length of the sequence):

• p r It increases the number with index p by r. (, )
  You have to output the number after the increase.

• p r It decreases the number with index p by r. (, ) You must not decrease the number if it would become negative.
  You have to output the number after the decrease.

s l r mod You have to output the sum of numbers in the interval which are equal mod (modulo m). () ()
Input
The first line of each test case contains the number of elements of the sequence n and the number m. (1 ≤ n ≤ 10000) (1 ≤ m ≤ 10)

The second line contains n initial numbers of the sequence. (0 ≤ number ≤ 1000000000)

The third line of each test case contains the number of queries q (1 ≤ q ≤ 10000).

The following q lines contains the queries (one query per line).

Output
Output q lines - the answers to the queries.

Examples
input
3 4
1 2 3
3
s 1 3 2
+ 2 1
- 1 2
output
2
3
1

#include
#define MAXN 10010
#define LL long long
using namespace std;
struct BIT {
    LL c[MAXN];
    int n;
    int lowbit(int x){ return x&(-x); }
    void modify(int x,LL y) {
        for(;x<=n;x+=lowbit(x)) c[x]+=y;
    }
    LL query(int x){
        LL ret=0;
        for(;x;x-=lowbit(x)) ret+=c[x];
        return ret;
    }
    LL query(int l,int r){ return query(r)-query(l-1); }
}bit[20];
int n,m,T;
LL a[MAXN];
int main() {
    scanf("%d%d",&n,&m);
    for(int i=0; i<m; i++) bit[i].n=n;
    for(int i=1; i<=n; i++) {
        scanf("%lld",&a[i]);
        bit[a[i]%m].modify(i,a[i]);
    }
    scanf("%d",&T);
    while(T--){
        int x,y,z;
        char opt[10];
        scanf("%s%d%d",opt,&x,&y);
        if(opt[0]=='+'){
            bit[a[x]%m].modify(x,-a[x]);
            a[x]+=y;
            bit[a[x]%m].modify(x,a[x]);
            printf("%lld\n",a[x]);
        }
        if(opt[0]=='-'){
            if(a[x]<y) printf("%lld\n",a[x]);
            else {
                bit[a[x]%m].modify(x,-a[x]);
                a[x]-=y;
                bit[a[x]%m].modify(x,a[x]);
                printf("%lld\n",a[x]);
            }
        }
        if(opt[0]=='s'){
            scanf("%d",&z);
            printf("%lld\n",bit[z].query(x,y));
        }
    }
    return 0;
}