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Codeforce GYM 100741 A. Queries

2018-06-17 22:03:44来源：未知阅读 ()

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. ( $\text{[math]}$ , $\text{[math]}$ )
You have to output the number after the increase.
- p r It decreases the number with index p by r. ( $\text{[math]}$ , $\text{[math]}$ ) 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 $\text{[math]}$ which are equal mod (modulo m). ( $\text{[math]}$ ) ( $\text{[math]}$ )

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

output

2
3
1


看一下数据范围，
m<=20
果断开20个树状数组
注意：要开long long

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstring>
 6 #include<algorithm>
 7 #define lli long long int 
 8 using namespace std;
 9 const lli MAXN=10001;
10 void read(lli &n)
11 {
12     char c='+';lli x=0;bool flag=0;
13     while(c<'0'||c>'9'){c=getchar();if(c=='-')flag=1;}
14     while(c>='0'&&c<='9')x=x*10+c-48,c=getchar();
15     n=flag==1?-x:x;
16 }
17 lli n,m;
18 struct node
19 {
20     lli a[MAXN];
21     lli lowbit(lli x){return x&(-x);}
22     lli change(lli pos,lli val)
23     {
24         for(;pos<=n;pos+=lowbit(pos))
25             a[pos]+=val;
26     }
27     lli query(lli l,lli r)
28     {
29         return ask(r)-ask(l-1);
30     }
31     lli ask(lli pos)
32     {
33         lli ans=0;
34         for(;pos;pos-=lowbit(pos))
35             ans+=a[pos];
36         return ans;
37     }    
38     node(){memset(a,0,sizeof(a));}
39 }bit[21];
40 lli date[MAXN];
41 int main()
42 {
43     read(n);read(m);
44     for(lli i=1;i<=n;i++)
45     {
46         read(date[i]);
47         bit[date[i]%m].change(i,date[i]);
48     }
49     lli T;read(T);
50     while(T--)
51     {
52         char how;cin>>how;
53         if(how=='+')
54         {
55             lli p,num;read(p);read(num);
56             bit[date[p]%m].change(p,-date[p]);
57             date[p]+=num;
58             bit[date[p]%m].change(p,date[p]);
59             printf("%lld\n",date[p]);
60         }
61         else if(how=='-')
62         {
63             lli p,num;read(p);read(num);
64             if(date[p]<num)
65                 printf("%lld\n",date[p]);
66             else
67             {
68                 bit[date[p]%m].change(p,-date[p]);
69                 date[p]-=num;
70                 bit[date[p]%m].change(p,date[p]);
71                 printf("%lld\n",date[p]);
72             }
73             
74         }
75         else if(how=='s')
76         {
77             lli l,r,mod;read(l);read(r);read(mod);
78             printf("%lld\n",bit[mod].query(l,r));
79         }
80     }
81     return 0;
82 }