HDU 4135 容斥原理
链接
题意:给你一个区间
,求这个区间内与n互质的数的个数。
题解:可以先求出
内与n互质的数的个数,然后再求出
内与n互质的数的个数,然后用前者减去后者即是答案。二者都可以采用容斥原理求。对容斥原理不了解的可以看链接。首先求出n的所有质因子,然后dfs求容斥系数。
代码:
#include<cstdio>
#include<algorithm>
#include<vector>
#include<cstring>
#include<cmath>
using namespace std;
const int mod = 200907;
typedef long long LL;
const int MAXN = 1e5+100;
LL prime[MAXN+10];
void getprime()
{
for(int i = 0; i < MAXN; i++) prime[i] = 0;
for(int i = 2; i <= MAXN; i++)
{
if(!prime[i]) prime[++prime[0]] = i;
for(int j = 1; j <= prime[0]&&prime[j]<=MAXN/i; j++)
{
prime[prime[j]*i] = 1;
if(i%prime[j]==0)break;
}
}
}
//分解因子
LL factor[100][2];
int fatcnt;
void getfactors(int x)
{
fatcnt = 0;
int tmp = x;
for(int i = 1; prime[i]<=tmp/prime[i];i++)
{
if(tmp%prime[i]==0)
{
factor[fatcnt][0] = prime[i];
while(tmp%prime[i] == 0)
{
factor[fatcnt][1]++;
tmp /= prime[i];
}
fatcnt++;
}
}
if(tmp!=1)
{
factor[fatcnt][0] = tmp;
factor[fatcnt][1] = 1;
fatcnt++;
}
}
//求容斥系数
LL fact[1000][2]; //第一列存容斥因子,第二列存符号
int facnt;
void dfs(LL cur, LL st,LL type)
{
if(st==fatcnt)
{
fact[facnt][0] = cur;
fact[facnt][1] = type;
facnt++;
return ;
}
dfs(cur,st+1,type);
dfs(cur*factor[st][0],st+1,-type);
}
LL solve(LL n)
{
LL ans = 0;
for(int i = 0;i<facnt;i++)
{
ans += fact[i][1]*n/fact[i][0];
}
return ans;
}
int main()
{
int t, cas = 0;
scanf("%d", &t);
getprime();
while(t--)
{
LL a,b,n;
scanf("%lld%lld%lld", &a, &b, &n);
memset(factor,0,sizeof(factor));
memset(fact,0,sizeof(fact));
facnt = 0;
getfactors(n);
dfs(1,0,1);
LL ans = solve(b)-solve(a-1);
printf("Case #%d: %lld\n", ++cas,ans);
}
return 0;
}