hdu 1695 容斥原理或莫比乌斯反演

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5053    Accepted Submission(s): 1812


[align=left]Problem Description[/align]
Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total
number of different number pairs.

Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

 

[align=left]Input[/align]
The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.

Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

 

[align=left]Output[/align]
For each test case, print the number of choices. Use the format in the example.

 

[align=left]Sample Input[/align]

2
1 3 1 5 1
1 11014 1 14409 9

 

[align=left]Sample Output[/align]

Case 1: 9
Case 2: 736427

HintFor the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).

一道比较经典的题目，可以用容斥原理，也可以莫比乌斯反演，莫比乌斯反演暂时没实现，就是把b，d都除以k，然后查找互素对数，枚举+容斥，一个特判的trick纠结了一天，

被dream神三分钟就找到了，

代码：

#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef __int64 ll;
bool isprime[1001000];
ll prime[1001000],factor[1010];
void getprime(){
memset(isprime,1,sizeof(isprime));isprime[1]=0;
ll &cnt=prime[0];cnt=0;
for(ll i=2;i<=1000000;i++){
if(isprime[i])prime[++cnt]=i;
for(ll j=1;j<=cnt&&i*prime[j]<=1000000;j++){
isprime[i*prime[j]]=0;
if(i%prime[j]==0)break;
}
}
// cout<<prime[0]<<endl;
}
void getfactor(ll x){
ll &cnt=factor[1000];cnt=0;
for(ll i=1;i<=prime[0]&&prime[i]*prime[i]<=x;i++)
if(x%prime[i]==0){
factor[cnt++]=prime[i];
while(x%prime[i]==0)x/=prime[i];
}
if(x>1)factor[cnt++]=x;
}
ll cal(ll x){
ll cnt=factor[1000],ans=0;
for(ll i=1;i<(1<<cnt);i++){
ll pp=1,ss=0;
for(ll j=0;j<cnt;j++)
if(i&(1<<j)){
ss++;pp*=factor[j];
}
if(ss%2)ans+=x/pp;
else ans-=x/pp;
}
return x-ans;
}
int main()
{
//freopen("data.in","r",stdin);
// freopen("data.out","w",stdout);
getprime();
int T;scanf("%d",&T);
for(int t=1;t<=T;t++){
ll a,b,c,d,k;
scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&k);
printf("Case %d: ",t);
if(k==0){
puts("0");continue;
}
ll ans=0;
if(b>d)swap(b,d);b/=k;d/=k;
if(b)ans+=d;
for(ll i=2;i<=b;i++){
getfactor(i);
ans+=cal(d)-cal(i);
}
cout<<ans<<endl;
}
return 0;
}