HDU 4947 GCD Array

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 descriptionteacher Mai finds that implements problems about arithmetic function can be implicitly CED 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

 

Inputthere 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 ).

 

Outputfor each case, output "case # K:" First, where k is the case number counting from 1.

Then output the answer to each query.

 

Sample input6 41 4 1 22 51 3 32 30 0

 

Sample output case #1: 67

 

Authorxudyh

 

Source2014 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 }

 

HDU 4947 GCD Array