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\n3 5 7\n");
}
}
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，\(c_i = \max\limits_{j=1}^{n}\{k_j\}\)
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 \(\lfloor x \rfloor\)
#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]," \n"[j==n]);
}
}
}
signed main(void) {
Init();
Work();
return 0;
}
