HDU 4947 GCD Array 容斥原理+树状数组
GCD Array
Time Limit: 11000/5500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 843 Accepted Submission(s):
205
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
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).
case number counting from 1.
Then output the answer to each query.
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef __int64 LL; const int maxn = +;
const int INF = 2e5+;
LL p[maxn];
bool s[INF];
int prime[],len; void Init()
{
len = ;
memset(s,false,sizeof(s));
for(int i=;i<INF;i++)
{
if(s[i]==true)continue;
prime[++len] = i;
for(int j=i+i;j<INF;j=j+i)
s[j]=true;
}
}
void add(int x,int n,int num1)
{
for(int i=x;i<=n;i=i+(i&(-i)))
p[i] = p[i] + num1;
}
LL query(int x)
{
if(x==)return ;
LL sum1 = ;
while(x)
{
sum1=sum1+p[x];
x=x-(x&(-x));
}
return sum1;
}
int Q[],yz[],ylen,qlen;
void init(int n)
{
ylen = qlen = ;
for(int i=;prime[i]*prime[i]<=n;i++)
{
if(n%prime[i]==)
{
while(n%prime[i]==) n=n/prime[i];
yz[++ylen] = prime[i];
}
}
if(n!=) yz[++ylen] = n;
Q[]=-;
for(int i=;i<=ylen;i++)
{
int k = qlen;
for(int j=;j<=k;j++)
Q[++qlen] = -*Q[j]*yz[i];
}
}
int main()
{
int n,m,hxl,d,v,size1,x,T=;
Init();
while(scanf("%d%d",&n,&m)>)
{
if(n==&&m==)break;
memset(p,,sizeof(p));
printf("Case #%d:\n",++T);
while(m--)
{
scanf("%d",&size1);
if(size1==)
{
scanf("%d%d%d",&hxl,&d,&v);
if(hxl%d!=)continue;
hxl = hxl /d;
int tom = n/d;
add(d,n,v);
init(hxl);
for(int i=;i<=qlen;i++)
if(Q[i]<) {
Q[i] = -Q[i];
if(Q[i]>tom)continue;
add(Q[i]*d,n,v);
}
else {
if(Q[i]>tom)continue;
add(Q[i]*d,n,-v);
}
}
else{
scanf("%d",&x);
LL sum1 = ;
for(int i=,la=;i<=x;i=la+){
la = x/(x/i);
sum1 = sum1 + (query(la)-query(i-))*(x/i);
}
printf("%I64d\n",sum1);
}
}
}
return ;
}
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