hdu 3939 Sticks and Right Triangle 求<=N的素勾股数的个数容斥原理

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本文探讨了如何在无限长的棍子上切割三个整数长度的子段以形成直角三角形的问题。通过数学算法确定了在特定长度限制下可以形成的直角三角形的数量。

不会做直接贴

Problem Description

We have a stick with infinite length. Now we want to cut 3 sub-sticks with length x, y, z which is not large than L to form a right triangle. With unknown reasons we assume that x, y, z are all integers and satisfy that x, y, z are all co-primed each other. We want to know how many right triangles are there exist under our constraints

Input

The first line of input is an integer T (T<=5) indicating the number of test cases.
Each case contains a single integer L (L<=1,000,000,000,000).

Output

For each test case output a single integer in one line, indicating the number of right triangles.

Sample Input

1
5

Sample Output

1
Hint
In our test case, we could find a right triangle (3,4,5) which satisfy 3,4,5<=5 and gcd(3,4)=1,gcd(3,5)=1,gcd(4,5)=1.

#include <iostream>
#include <fstream>
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
#define MAXN 1000000

using namespace std;

int pri[MAXN+10],pri_num,tab_phi[MAXN+10];
bool temp[MAXN+10];
int T,num,Dec[20];
long long ans,n;

void print_prime()
{
pri_num=0;
for(int i=2;i<MAXN;i+=2) temp[i]=1;
pri[pri_num++]=2;
tab_phi[2]=1;
tab_phi[1]=1;
for(int i=3;i<MAXN;i+=2)
{
if(!temp[i]) pri[pri_num++]=i,tab_phi[i]=i-1;
for(int j=0;j<pri_num&&i*pri[j]<MAXN;j++)
{
temp[i*pri[j]]=1;
if(i%pri[j]==0) {tab_phi[i*pri[j]]=tab_phi[i]*pri[j];break;}
else tab_phi[i*pri[j]]=tab_phi[i]*(pri[j]-1);
}
if((((i+1)>>1)&1)==0) tab_phi[i+1]=tab_phi[(i+1)>>1]<<1;
else tab_phi[i+1]=tab_phi[(i+1)>>1];
}
}

void prime_dec(int a)
{
num=0;
if(!temp[a]) {Dec[num++]=a; return;}
for(int i=0;i<pri_num&&a>1;i++)
{
if(a%pri[i]==0)
{
Dec[num++]=pri[i];
while(a%pri[i]==0)
{
a/=pri[i];
}
if(a>1&&(!temp[a])) {Dec[num++]=a; return;}
}
}
}

int dfs(int k,int l,int s,int a)
{
if(k==num)
{
if(l&1) ans-=a/s;
else ans+=a/s;
return 0;
}
dfs(k+1,l,s,a);
dfs(k+1,l+1,s*Dec[k],a);
return 0;
}

int main()
{
print_prime();
cin>>T;
while(T--)
{
ans=0;
scanf("%I64d",&n);
int m=(int)sqrt(n+0.0);
for(int i=m;i>0;i--)
{
int p=(int)sqrt(n-(long long)i*i+0.0);
if(i&1)
{
prime_dec(i);
if(i<=p) dfs(0,0,1,i>>1);
else dfs(0,0,1,p>>1);
}
else
{
if(i<=p) ans+=tab_phi[i];
else { prime_dec(i); dfs(0,0,1,p); }
}
}
printf("%I64d\n",ans);
}
return 0;
}