codeforce GYM 100741 A Queries

A. Queries

time limit per test:0.25 s

memory limit per test:64 MB

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

题意

一个长度为 n 的序列，q 次三种操作，

+ p r: 下标为 p 的数加 r.
- p r: 下表为 p 的数减 r.
s l r mod: 询问在区间[l,r]中模 m 等于 mod 的所有数的和。

分析

可以建立m个树状数组，然后询问就好处理了，加减要现在原来的树状数组中减掉，然后在之后的树状数组中加上。

code

 #include<cstdio>
 #include<algorithm>

 using namespace std;
 typedef long long LL;

 const int MAXN = ;
 LL n,m,q;
 LL s[MAXN];

 LL read()
 {
     LL x = ,f = ;char ch = getchar();
     while (ch<''||ch>'') {if (ch=='-') f=-; ch = getchar(); }
     while (ch>=''&&ch<='') {x = x*+ch-''; ch = getchar(); }
     return x*f;
 }

 struct Tree_array{
     LL a[MAXN];
     int lowbit(int x)
     {
         return x&(-x);
     }
     void update(int x,int v)
     {
         for (; x<=n; x+=lowbit(x)) a[x] += v;
     }
     LL query(int x)
     {
         LL ret = ;
         for (; x; x-=lowbit(x))
          ret += a[x];
         return ret;
     }
 }t[];

 int main()
 {
     n = read();m = read();
     for (int i=; i<=n; ++i)
     {
         s[i] = read();
         t[s[i]%m].update(i,s[i]);
     }
     char opt[];
     q = read();
     LL x,y,mo;
     while (q--)
     {
         scanf("%s",opt);
         x = read();y = read();
         if (opt[]=='s')
         {
             mo = read();
             printf("%lld\n",t[mo].query(y)-t[mo].query(x-));
         }
         else if (opt[]=='+')
         {
             t[s[x]%m].update(x,-s[x]);
             s[x] += y;
             t[s[x]%m].update(x,s[x]);
             printf("%lld\n",s[x]);
         }
         else
         {
             if (s[x]<y) printf("%lld\n",s[x]);
             else
             {
                 t[s[x]%m].update(x,-s[x]);
                 s[x] -= y;
                 t[s[x]%m].update(x,s[x]);
                 printf("%lld\n",s[x]);
             }
         }
     }
     return ;
 }