源出处:https://blog.csdn.net/qingshui23/article/details/51077728
POJ 3904 Sky Code (容斥原理)
2016年04月06日 19:53:49
阅读数：827
Sky Code
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 2027 | Accepted: 650 |
Description
Stancu likes space travels but he is a poor software developer and will never be able to buy his own spacecraft. That is why he is preparing to steal the spacecraft of Petru. There is only one problem – Petru has locked the spacecraft with a sophisticated cryptosystem based on the ID numbers of the stars from the Milky Way Galaxy. For breaking the system Stancu has to check each subset of four stars such that the only common divisor of their numbers is 1. Nasty, isn’t it? Fortunately, Stancu has succeeded to limit the number of the interesting stars to N but, any way, the possible subsets of four stars can be too many. Help him to find their number and to decide if there is a chance to break the system.
Input
In the input file several test cases are given. For each test case on the first line the number N of interesting stars is given (1 ≤ N ≤ 10000). The second line of the test case contains the list of ID numbers of the interesting stars, separated by spaces. Each ID is a positive integer which is no greater than 10000. The input data terminate with the end of file.
Output
For each test case the program should print one line with the number of subsets with the asked property.
Sample Input
4 2 3 4 5 4 2 4 6 8 7 2 3 4 5 7 6 8
Sample Output
1 0 34
Source
Southeastern European Regional Programming Contest 2008
题目大意：
给出n(n<10000)个数，这些数<=10000，让你求的就是 GCD(a,b,c,d) == 1的个数，注意必须是 4 个数
解题思路：
其实这个题一开始做就像暴力，可是自己又暴不明白，因为 他把这个题归入容斥原理，所以就考虑GCD(a,b,c,d) == 1的反面就是 GCD(a,b,c,d) != 1，那么a, b, c, d 必定有同样的因子，所以 ：
第一步：计算n个数字选四个总共有多少种选法[ C(n,4) ]，然后减去公约数不是1的。
第二步：由一个素因子组合成的四个数的组合数 + 由两个素因子组合成的四个数的组合数 -....+..(奇加偶减)
第三步：进行二进制枚举，在枚举的时候要用到两个数组 ，_cnt[]：记录当前因子是由多少个素因子组成, sum[]：记录当前因子的个数
第四步：我们先考虑将每一个数 进行素因子分解，在这里完全可以采用暴力也就是 O(sqrt(n))的算法，因为数据不是很大，当然也可以直接进行素数筛，我采用的就是素数筛。
最后一步：假设a是2的因子的个数，b是3的因子的个数，c是6的因子的个数，就是C(n,4) - C(a,4) - C(b,4) + C(c,4)
容斥原理，与第一步差不多...
PS:我还用了输入输出外挂，果然快呀 ^_^
My Code：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int MAXN = 1e4+5;
typedef long long LL;
int Scan()///输入外挂
{
int res=0,ch,flag=0;
if((ch=getchar())=='-')
flag=1;
else if(ch>='0'&&ch<='9')
res=ch-'0';
while((ch=getchar())>='0'&&ch<='9')
res=res*10+ch-'0';
return flag?-res:res;
}
void Out(LL a)///输出外挂
{
if(a>9)
Out(a/10);
putchar(a%10+'0');
}
int p[MAXN];
bool prime[MAXN];
int k;
void isprime()///素数筛
{
k = 0;
LL i, j;
memset(prime,false,sizeof(prime));
for(i=2; i<MAXN; i++)
{
if(!prime[i])
{
p[k++] = i;
for(j=i*i; j<MAXN; j+=i)
prime[j] = true;
}
}
}
int fac[MAXN/100],num[100], cnt;
void Dec(int m)///素因子分解
{
cnt = 0;
memset(num, 0, sizeof(num));
for(int i=0; p[i]*p[i]<=m&&i<k; i++)
{
if(m%p[i]==0)
{
fac[cnt] = p[i];
while(m%p[i]==0)
{num[cnt]++;
m /= p[i];
}
cnt++;
}
}
if(m > 1)
{
fac[cnt] = m;
num[cnt++]=1;
}
///cout<<cnt<<endl;
///cout<<fac[0]<<" "<<fac[1]<<endl;
}
int sum[MAXN];///记录当前因子的个数
int _cnt[MAXN];///记录当前因子是由多少个素因子组成
void Get(int m)
{
Dec(m);
for(int i=1; i<(1<<cnt); i++)
{
int s = 0;
LL ans = 1;
for(int j=0; j<cnt; j++)
{
if(i & (1<<j))
{
s++;
ans *= fac[j];
if(ans > m)
break;
}
}
sum[ans]++;
_cnt[ans] = s;
}
}
LL ak[MAXN];///C(n,4)
void RongChi()
{
memset(ak,0,sizeof(ak));
for(LL i=4; i<MAXN; i++)
ak[i] = i*(i-1)*(i-2)*(i-3)/24;
}
int main()
{
RongChi();
isprime();
///Dec(1);
int n, x;
while(~scanf("%d",&n))
{
memset(sum, 0, sizeof(sum));
int m = n;
while(m--)
{
x = Scan();
Get(x);
}
LL ret = 0;
for(int i=0; i<MAXN; i++)
{
if(sum[i])
{
if(_cnt[i] & 1)
ret += ak[sum[i]];
else
ret -= ak[sum[i]];
}
}
LL ans = ak[n] - ret;
Out(ans);
puts("");
}
return 0;
}