证明见luogu,我是打表发现这个规律的。但是这个规律也很显然吧。任何大于5的自然数都可以拆成若干个2,3的和,拆成2,3再乘肯定更大
namespace solve {
int num2, num3;
const int Len = 1007, W = 1000000000;
struct hp {
lxl x[Len];
int siz;
inline void trim() {
while(siz && !x[siz - 1])--siz;
}
hp(lxl a = 0) {
memset(x, 0, sizeof x);
siz = 0;
while(a) {
x[siz++] = a % W;
a /= W;
}
}
} ans, p, q, res;
inline void operator +=(hp &a, const hp &b) {
a.siz = max(a.siz, b.siz) + 1;
rep(i, 0, a.siz - 1) {
a.x[i] += b.x[i];
}
rep(i, 0, a.siz - 1) {
if(a.x[i] >= W) {
++a.x[i + 1];
a.x[i] -= W;
}
}
a.trim();
}
inline hp operator +(hp a, const hp &b) {
a += b;
return a;
}
inline hp operator *(const hp &a, const hp &b) {
hp c;
c.siz = b.siz + a.siz;
rep(i, 0, a.siz - 1) {
rep(j, 0, b.siz - 1) {
int k = i + j;
c.x[k] += a.x[i] * b.x[j];
c.x[k + 1] += c.x[k] / W;
c.x[k] %= W;
}
}
c.trim();
return c;
}
inline int get(lxl x) {
int num(0);
while(x) {
num++;
x /= 10;
}
return num;
}
inline std::string tr(lxl x) {
std::string s = "";
while(x) {
s += x % 10 + '0';
x /= 10;
}
reverse(s.begin(), s.end());
return s;
}
inline void Write(const hp &a) {
int num(0);
lxl r(a.x[a.siz - 1]);
num = get(r);
cout << (a.siz - 1) * 9 + num << '
';
std::string s = "";
s += tr(a.x[a.siz - 1]);
drp(i, a.siz - 2, 0) {
num = get(a.x[i]);
while(num < 9) {
s += '0';
++num;
}
s += tr(a.x[i]);
}
int t = s.size() - 1;
rep(i, 0, min(99, t))
cout << s[i];
cout << '
';
}
int main() {
if(n == 1 || n == 2) {
cout << 1 << '
' << n << '
';
return 0;
}
int t = n % 3;
if(t == 1) {
num2 = 2;
} else if(t == 2) {
num2 = 1;
} else {
num2 = 0;
}
num3 = (n - num2 * 2) / 3;
p = 2;
q = 3;
res = 1;
ans = 1;
if(num2) {
while(num2 > 1) {
p = p * 2;
--num2;
}
ans = ans * p;
}
if(num3) {
while(num3 > 1) {
q = q * 3;
--num3;
}
ans = ans * q;
}
Write(ans);
return 0;
}
}