CF717B
算法1
dp!
时间复杂度
算法二
贪心！
显然这题就是求一棵二叉树，每个点到左儿子距离为
扩展
时间复杂度
算法三
类似于NOIp2016蚯蚓用3个队列优化一发。
时间复杂度
算法四
原题出题人的算法。
用map一次扩展多个距离相同的点。
能AC但是是假算法出题人没卡。
时间复杂度
算法五
我们二分最后一次扩展的点离原点的距离w。
check的时候，考虑所有
设
然后就可以枚举
最后计算答案的时候，考虑一次扩展对答案的贡献：
首先计算
非常优秀。
/*
Author: CNYALI_LK
LANG: C++
PROG: B.cpp
Mail: cnyalilk@vip.qq.com
*/
#include<bits/stdc++.h>
#define debug(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG printf("Passing [%s] in LINE %lld\n",__FUNCTION__,__LINE__)
#define Debug debug("Passing [%s] in LINE %lld\n",__FUNCTION__,__LINE__)
#define all(x) x.begin(),x.end()
#define x first
#define y second
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
const signed inf=0x3f3f3f3f;
const double eps=1e-8;
const double pi=acos(-1.0);
template<class T>ll chkmin(T &a,T b){return a>b?a=b,1:0;}
template<class T>ll chkmax(T &a,T b){return a<b?a=b,1:0;}
template<class T>T sqr(T a){return a*a;}
template<class T>T mmin(T a,T b){return a<b?a:b;}
template<class T>T mmax(T a,T b){return a>b?a:b;}
template<class T>T aabs(T a){return a<0?-a:a;}
template<class T>ll dcmp(T a,T b){return a>b;}
template<ll *a>ll cmp_a(ll x,ll y){return a[x]<a[y];}
#define min mmin
#define max mmax
#define abs aabs
namespace io {
const ll SIZE = (1 << 21) + 1;
char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; ll f, qr;
// getchar
#define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
// prll the remaining part
inline void flush () {
fwrite (obuf, 1, oS - obuf, stdout);
oS = obuf;
}
// putchar
inline void putc (char x) {
*oS ++ = x;
if (oS == oT) flush ();
}
// input a signed lleger
inline void read (signed &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}
inline void read (long long &x) {
for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
}
inline void read (char &x) {
x=gc();
}
inline void read(char *x){
while((*x=gc())=='\n' || *x==' '||*x=='\r');
while(!(*x=='\n'||*x==' '||*x=='\r'))*(++x)=gc();
}
template<typename A,typename ...B>
inline void read(A &x,B &...y){
read(x);read(y...);
}
// prll a signed lleger
inline void write (signed x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
}
inline void write (long long x) {
if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
while (x) qu[++ qr] = x % 10 + '0', x /= 10;
while (qr) putc (qu[qr --]);
}
inline void write (char x) {
putc(x);
}
inline void write(const char *x){
while(*x){putc(*x);++x;}
}
inline void write(char *x){
while(*x){putc(*x);++x;}
}
template<typename A,typename ...B>
inline void write(A x,B ...y){
write(x);write(y...);
}
//no need to call flush at the end manually!
struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: read;
using io :: putc;
using io :: write;
ll c0,c1,n;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lls[31];
ll C(ll a,ll b){
ll s=1,x,g;
for(ll i=1;i<=b;++i)lls[i]=i;
for(ll i=0;i<b;++i){
x=a+b-i;
for(ll j=1;j<=b;++j){
g=gcd(lls[j],x);
x/=g;
lls[j]/=g;
}
if(x>inf/s)return inf;
s*=x;
}
return s;
}
#define chk if(tmp>n)return 1
ll check(ll x){
ll tmp=x/c0+x/c1+1;
chk;
for(ll i=1;i<=26;++i){
for(ll j=i;j*c0+i*c1<=x;++j){
tmp+=C(j,i);
chk;
}
for(ll j=i+1;j*c1+i*c0<=x;++j){
tmp+=C(j,i);
chk;
}
}
return 0;
}
void Work(ll x,ll &tmp,ll &ans){
tmp=x/c0+x/c1+1;
ans=((x/c0)*(x/c0+1)*c0+(x/c1)*(x/c1+1)*c1)>>1;
ll c;
for(ll i=1;i<=26;++i){
for(ll j=i;j*c0+i*c1<=x;++j){
c=C(j,i);
tmp+=c;
ans+=c*(j*c0+i*c1);
}
for(ll j=i+1;j*c1+i*c0<=x;++j){
c=C(j,i);
tmp+=c;
ans+=c*(j*c1+i*c0);
}
}
}
int main(){
#ifdef cnyali_lk
freopen("B.in","r",stdin);
freopen("B.out","w",stdout);
#endif
read(n,c0,c1);
--n;
if(c0>c1)swap(c0,c1);
if(!c0)printf("%lld\n",n*c1);
ll l=0,r=c1*26,mid;
while(l<=r){
mid=(l+r)>>1;
if(check(mid))r=mid-1;
else l=mid+1;
}
ll c,s;
Work(l-1,c,s);
s+=(n-c)*l+n*(c0+c1);
printf("%lld\n",s);
return 0;
}