Codeforces GYM 100741A . Queries

time limit per test

0.25 seconds

memory limit per test

64 megabytes

input

standard input

output

standard 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

题目大意：
s 求 l~r中 对m取模==mod 的 和
+ 单点修改 
- 单点修改,如果减后小于0直接输出 

树状数组 
屠龙宝刀点击就送

#include <ctype.h>
#include <cstdio>
#define N 10005
typedef long long LL;
void read(LL &x)
{
    x=0;bool f=0;
    char ch=getchar();
    for(;!isdigit(ch);ch=getchar()) if(ch=='-') f=1;
    for(;isdigit(ch);ch=getchar()) x=x*10+ch-'0';
    x=f?(~x)+1:x;
}
LL dis[N],n,m,q;
struct node
{
    LL tag[N];
    int n;
    int lowbit(int x) {return x&((~x)+1);}
    void plus(int x,int y)
    {
        for(;x<=n;x+=lowbit(x)) tag[x]+=y;
    }
    LL query(int x)
    {
        LL ans=0;
        for(;x;x-=lowbit(x)) ans+=tag[x];
        return ans;
    }
}a[25];
int main()
{
    read(n);read(m);
    for(int i=0;i<m;i++) a[i].n=n;
    for(int i=1;i<=n;i++)
    {
        read(dis[i]);
        a[dis[i]%m].plus(i,dis[i]);
    }
    char str[5];
    read(q);
    for(LL x,y,z;q--;)
    {
        scanf("%s",str+1);read(x);read(y);
        switch(str[1])
        {
            case 's':
            {
                read(z);
                printf("%lld\n",a[z].query(y)-a[z].query(x-1));
                break;
            }
            case '+':
            {
                a[dis[x]%m].plus(x,-dis[x]);
                dis[x]+=y;
                a[dis[x]%m].plus(x,dis[x]);
                printf("%lld\n",dis[x]);
                break;
            }
            case '-':
            {
                if(dis[x]<y) {printf("%lld\n",dis[x]);}
                else 
                {
                    a[dis[x]%m].plus(x,-dis[x]);
                    dis[x]-=y;
                    a[dis[x]%m].plus(x,dis[x]);
                    printf("%lld\n",dis[x]);
                }
                break;
            }
        }
    }
    return 0;
}