GCD Array
Time Limit: 11000/5500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 843 Accepted Submission(s):
205
Problem Description
Teacher Mai finds that many problems about arithmetic
function can be reduced to the following problem:
Maintain an array a
with index from 1 to l. There are two kinds of operations:
1. Add v to
ax for every x that gcd(x,n)=d.
2. Query 
Input
There are multiple test cases, terminated by a line "0
0".
For each test case, the first line contains two integers
l,Q(1<=l,Q<=5*10^4), indicating the length of the array and the number of
the operations.
In following Q lines, each line indicates an operation,
and the format is "1 n d v" or "2 x"
(1<=n,d,v<=2*10^5,1<=x<=l).
Output
For each case, output "Case #k:" first, where k is the
case number counting from 1.
Then output the answer to each query.
Sample Input
6 4
1 4 1 2
2 5
1 3 3 3
2 3
0 0
Sample Output
Case #1:
6
7
Author
xudyh
Source
2014
Multi-University Training Contest 8
1 #include<iostream>
2 #include<stdio.h>
3 #include<cstring>
4 #include<cstdlib>
5 using namespace std;
6 typedef __int64 LL;
7
8 const int maxn = 50000+3;
9 const int INF = 2e5+3;
10 LL p[maxn];
11 bool s[INF];
12 int prime[17985],len;
13
14 void Init()
15 {
16 len = 0;
17 memset(s,false,sizeof(s));
18 for(int i=2;i<INF;i++)
19 {
20 if(s[i]==true)continue;
21 prime[++len] = i;
22 for(int j=i+i;j<INF;j=j+i)
23 s[j]=true;
24 }
25 }
26 void add(int x,int n,int num1)
27 {
28 for(int i=x;i<=n;i=i+(i&(-i)))
29 p[i] = p[i] + num1;
30 }
31 LL query(int x)
32 {
33 if(x==0)return 0;
34 LL sum1 = 0;
35 while(x)
36 {
37 sum1=sum1+p[x];
38 x=x-(x&(-x));
39 }
40 return sum1;
41 }
42 int Q[5003],yz[1000],ylen,qlen;
43 void init(int n)
44 {
45 ylen = qlen = 0;
46 for(int i=1;prime[i]*prime[i]<=n;i++)
47 {
48 if(n%prime[i]==0)
49 {
50 while(n%prime[i]==0) n=n/prime[i];
51 yz[++ylen] = prime[i];
52 }
53 }
54 if(n!=1) yz[++ylen] = n;
55 Q[0]=-1;
56 for(int i=1;i<=ylen;i++)
57 {
58 int k = qlen;
59 for(int j=0;j<=k;j++)
60 Q[++qlen] = -1*Q[j]*yz[i];
61 }
62 }
63 int main()
64 {
65 int n,m,hxl,d,v,size1,x,T=0;
66 Init();
67 while(scanf("%d%d",&n,&m)>0)
68 {
69 if(n==0&&m==0)break;
70 memset(p,0,sizeof(p));
71 printf("Case #%d:\n",++T);
72 while(m--)
73 {
74 scanf("%d",&size1);
75 if(size1==1)
76 {
77 scanf("%d%d%d",&hxl,&d,&v);
78 if(hxl%d!=0)continue;
79 hxl = hxl /d;
80 int tom = n/d;
81 add(d,n,v);
82 init(hxl);
83 for(int i=1;i<=qlen;i++)
84 if(Q[i]<0) {
85 Q[i] = -Q[i];
86 if(Q[i]>tom)continue;
87 add(Q[i]*d,n,v);
88 }
89 else {
90 if(Q[i]>tom)continue;
91 add(Q[i]*d,n,-v);
92 }
93 }
94 else{
95 scanf("%d",&x);
96 LL sum1 = 0;
97 for(int i=1,la=0;i<=x;i=la+1){
98 la = x/(x/i);
99 sum1 = sum1 + (query(la)-query(i-1))*(x/i);
100 }
101 printf("%I64d\n",sum1);
102 }
103 }
104 }
105 return 0;
106 }