Codeforce GYM 100741 A. Queries
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 2output
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 }