牛客挑战赛44

A

设三个数分别为 (k, k + 2, k + 4)

发现三个数在模 (3) 意义下构成了整个剩余系

所以只有当 (k=3) 为质数时有答案，也就是满足题目要求的数对只有 ((3,5,7))

#include<bits/stdc++.h>
using namespace std;

template < typename Tp >
inline void read(Tp &x) {
	x = 0; int fh = 1; char ch = 1;
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') fh = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
	x *= fh;
}

template < typename Tp >
inline void biread(Tp &x) {
	x = 0; int fh = 1; char ch = 1;
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') fh = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') x = x * 2 + ch - '0', ch = getchar();
	x *= fh;
}

long long n;

inline void Init(void) {
	cin >> n;
}

inline void Work(void) {
	if(n < 7) {
		puts("0");
	}
	else {
		printf("1
3 5 7
");
	}
}

signed main(void) {
	Init();
	Work();
	return 0;
}

---

C

(forall x < p^k)， (x) 的质因数中 (p) 不会超过 (k) 个，上述命题很容易证明。

考虑 (operatorname{lcm} {1,2, cdots,n } = prod{p_i^{c_i}})，其中，(c_i = maxlimits_{j=1}^{n}{k_j})

因此，对 LCM 有贡献的数，一定为 (p^k) ，只有在质数的若干次方，才会使对应的 (c_i) 变大。

问题转化为求 ([1,n]) 内，有多少个数可以表示成 (p^k) 的形式。

百度搜索 n 以内素数个数 1e11 会收获惊喜。

然后暴力就行了。

#include<cstdio>  
#include<cmath>  
#include<iostream>
using namespace std;  
#define LL long long  
const int N = 5e6 + 2;  
bool np[N];  
int prime[N], pi[N];  
int cnt;
int getprime()  
{  
    cnt = 0;  
    np[0] = np[1] = true;  
    pi[0] = pi[1] = 0;  
    for(int i = 2; i < N; ++i)  
    {  
        if(!np[i]) prime[++cnt] = i;  
        pi[i] = cnt;  
        for(int j = 1; j <= cnt && i * prime[j] < N; ++j)  
        {  
            np[i * prime[j]] = true;  
            if(i % prime[j] == 0)   break;  
        }  
    }  
    return cnt;  
}  
const int M = 7;  
const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;  
int phi[PM + 1][M + 1], sz[M + 1];  
void init()  
{  
    getprime();  
    sz[0] = 1;  
    for(int i = 0; i <= PM; ++i)  phi[i][0] = i;  
    for(int i = 1; i <= M; ++i)  
    {  
        sz[i] = prime[i] * sz[i - 1];  
        for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1];  
    }  
}  
int sqrt2(LL x)  
{  
    LL r = (LL)sqrt(x - 0.1);  
    while(r * r <= x)   ++r;  
    return int(r - 1);  
}  
int sqrt3(LL x)  
{  
    LL r = (LL)cbrt(x - 0.1);  
    while(r * r * r <= x)   ++r;  
    return int(r - 1);  
}  
LL getphi(LL x, int s)  
{  
    if(s == 0)  return x;  
    if(s <= M)  return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s];  
    if(x <= prime[s]*prime[s])   return pi[x] - s + 1;  
    if(x <= prime[s]*prime[s]*prime[s] && x < N)  
    {  
        int s2x = pi[sqrt2(x)];  
        LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2;  
        for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]];  
        return ans;  
    }  
    return getphi(x, s - 1) - getphi(x / prime[s], s - 1);  
}  
LL getpi(LL x)  
{  
    if(x < N)   return pi[x];  
    LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1;  
    for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1;  
    return ans;  
}  
LL lehmer_pi(LL x)  
{  
    if(x < N)   return pi[x];  
    int a = (int)lehmer_pi(sqrt2(sqrt2(x)));  
    int b = (int)lehmer_pi(sqrt2(x));  
    int c = (int)lehmer_pi(sqrt3(x));  
    LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2;  
    for (int i = a + 1; i <= b; i++)  
    {  
        LL w = x / prime[i];  
        sum -= lehmer_pi(w);  
        if (i > c) continue;  
        LL lim = lehmer_pi(sqrt2(w));  
        for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1);  
    }  
    return sum;  
}  
LL ans;
int main()   
{  
	init();
	LL n;
    cin>>n;
    ans+=lehmer_pi(n);
    for(int i=1;i<=cnt;++i){
    	LL x=prime[i];
    	for(LL tmp=x;tmp<=n;tmp*=x)if(tmp!=x)++ans;
	}
	cout<<ans<<endl;
    
    return 0;  
}  

---

E

2019年上海市高三数学竞赛第12题原题，百度即可。

其中，([x]) 是指 (lfloor x floor)

#include<bits/stdc++.h>
using namespace std;

template < typename Tp >
inline void read(Tp &x) {
	x = 0; int fh = 1; char ch = 1;
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') fh = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
	x *= fh;
}

template < typename Tp >
inline void biread(Tp &x) {
	x = 0; int fh = 1; char ch = 1;
	while(ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
	if(ch == '-') fh = -1, ch = getchar();
	while(ch >= '0' && ch <= '9') x = x * 2 + ch - '0', ch = getchar();
	x *= fh;
}
int n;
inline void Init(void) {
	read(n);
}
#define maxn 1005
int a[maxn][maxn];

inline void Work(void) {
	--n;
	int ans=n*n+2*n-1;
	ans>>=1;
	++n;
	cout<<ans<<endl;
	if(n%2==0){
		int tot=n*n+1;
		for(int i=1;i<=n;++i)a[i][1]=--tot;
		for(int i=(n>>1);i;--i){
			int t1=(i<<1),t2=t1-1;
			for(int j=2;j<=n;++j)a[t1][j]=--tot;
			for(int j=n;j>=2;--j)a[t2][j]=--tot;
		}
	}
	else{
		int tot=n*n+1;
		for(int i=1;i<=n;++i)a[i][1]=--tot;
		for(int i=n;i>3;i-=2){
			int t1=i,t2=i-1;
			for(int j=2;j<=n;++j)a[t1][j]=--tot;
			for(int j=n;j>=2;--j)a[t2][j]=--tot;
		}
		for(int j=2;j<=n;++j)a[3][j]=--tot;
		for(int j=n;j>1;j-=2){
			a[2][j]=--tot;
			a[1][j]=--tot;
			a[1][j-1]=--tot;
			a[2][j-1]=--tot;
		}
	}
	for(int i=1;i<=n;++i){
		for(int j=1;j<=n;++j){
			printf("%d%c",a[i][j]," 
"[j==n]);
		}
	}
}

signed main(void) {
	Init();
	Work();
	return 0;
}