Gym-101490H Multiplying Digits(进制转化--倒着)

传送门

题目说的是给一个数n，求在B进制下的每一位的乘积等于n的条件下的最小的数

先得到n小于b的因子，然后在因子乘积等于n的约束下组合因子，得到最小的mp[1]

保证最后的解，位数一定最少，而且不严格递增。

//china no.1#pragma comment(linker, "/STACK:1024000000,1024000000")#include <vector>#include <iostream>#include <string>#include <map>#include <stack>#include <cstring>#include <queue>#include <list>#include <stdio.h>#include <set>#include <algorithm>#include <cstdlib>#include <cmath>#include <iomanip>#include <cctype>#include <sstream>#include <functional>#include <stdlib.h>#include <time.h>#include <bitset>using namespace std;#define pi acos(-1)#define s_1(x) scanf("%d",&x)#define s_2(x,y) scanf("%d%d",&x,&y)#define s_3(x,y,z) scanf("%d%d%d",&x,&y,&z)#define s_4(x,y,z,X) scanf("%d%d%d%d",&x,&y,&z,&X)#define S_1(x) scan_d(x)#define S_2(x,y) scan_d(x),scan_d(y)#define S_3(x,y,z) scan_d(x),scan_d(y),scan_d(z)#define PI acos(-1)#define endl '\n'#define srand() srand(time(0));#define me(x,y) memset(x,y,sizeof(x));#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)#define close() ios::sync_with_stdio(0); cin.tie(0);#define FOR(x,n,i) for(int i=x;i<=n;i++)#define FOr(x,n,i) for(int i=x;i<n;i++)#define fOR(n,x,i) for(int i=n;i>=x;i--)#define fOr(n,x,i) for(int i=n;i>x;i--)#define W while#define sgn(x) ((x) < 0 ? -1 : (x) > 0)#define bug printf("***********\n");#define db double#define ll long long#define mp make_pair#define pb push_backtypedef long long LL;typedef pair <int, int> ii;const int INF=0x3f3f3f3f;const LL LINF=0x3f3f3f3f3f3f3f3fLL;const int dx[]={-1,0,1,0,1,-1,-1,1};const int dy[]={0,1,0,-1,-1,1,-1,1};const int maxn=4e3+10;const int maxx=4e5+10;const double EPS=1e-8;const double eps=1e-8;const int mod=1e9+7;template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}template <class T>inline bool scan_d(T &ret){char c;int sgn;if (c = getchar(), c == EOF){return 0;}while (c != '-' && (c < '0' || c > '9')){c = getchar();}sgn = (c == '-') ? -1 : 1;ret = (c == '-') ? 0 : (c - '0');while (c = getchar(), c >= '0' && c <= '9'){ret = ret * 10 + (c - '0');}ret *= sgn;return 1;}inline bool scan_lf(double &num){char in;double Dec=0.1;bool IsN=false,IsD=false;in=getchar();if(in==EOF) return false;while(in!='-'&&in!='.'&&(in<'0'||in>'9'))in=getchar();if(in=='-'){IsN=true;num=0;}else if(in=='.'){IsD=true;num=0;}else num=in-'0';if(!IsD){while(in=getchar(),in>='0'&&in<='9'){num*=10;num+=in-'0';}}if(in!='.'){if(IsN) num=-num;return true;}else{while(in=getchar(),in>='0'&&in<='9'){num+=Dec*(in-'0');Dec*=0.1;}}if(IsN) num=-num;return true;}void Out(LL a){if(a < 0) { putchar('-'); a = -a; }if(a >= 10) Out(a / 10);putchar(a % 10 + '0');}void print(LL a){ Out(a),puts("");}//freopen( "in.txt" , "r" , stdin );//freopen( "data.txt" , "w" , stdout );//cerr << "run time is " << clock() << endl;unsigned long long maxc=1ull<<63;map<LL,LL>Q;map<LL,LL>::reverse_iterator it;LL a[maxx];LL b,n;void solve(){    W(scanf("%I64d%I64d",&b,&n)!=EOF)    {        if(n==1)        {            print(1);            continue;        }        LL tmp=n,tot=0;        //首先对n进行分解        for(int i=2;i<b;i++)        {            if(n%i==0)  a[tot++]=i;            W(tmp%i==0)            {                tmp/=i;            }        }        if(tmp!=1)        {            puts("impossible");            continue;        }        Q.clear();        Q[n]=0;        for(it=Q.rbegin();it!=Q.rend();it++)        {            LL u=it->first,v=it->second;            LL mo=v%b;            for(int i=tot-1;i>=0;i--)            {                if(a[i]<mo) break;/*最优解的组成为非严格递增，所以若后一位小于前一位，则一定不是整体最优解*/                if(((long double)v*b+a[i])>maxc) continue;//防止爆LL                if(u%a[i]) continue;                LL x=u/a[i],y=v*b+a[i];                if(!Q[x]||Q[x]>y) Q[x]=y;            }        }        print(Q[1]);    }}int main(){    //freopen( "in.txt" , "r" , stdin );    //freopen( "data.txt" , "w" , stdout );    int t=1;    //s_1(t);    for(int cas=1;cas<=t;cas++)    {        //printf("Case %d:\n",cas);        solve();    }}