Gym 101612L Little Difference 因子分解

https://odzkskevi.qnssl.com/76ee6a37a5e7b6e0a1da0f46372ab4da?v=1512194397

题意：给你一个n，分解这个n，然后得到他的因子乘积，并且前后因子的差距最多为1。

比如可以 2 3 。 2 2。 但不能2 3 4。

做法：首先考虑如何得到这个n的因子x，一定是要靠近这个x也就是 x-1 x x+1，这三种与x接近的才能构成答案。

一直觉得是根号n的时间处理这个问题，但是n是1e18，后来看了别人的代码，用了二分的方法进行找因子，

具体是枚举幂项，二分1-n，找到mid^j是n的因子，腻害腻害，确实不会这种操作。最后就是要么全是mid^j构成n，

或者mid^j*(mid+1)^k构成n，记录答案即可。

///                 .-~~~~~~~~~-._       _.-~~~~~~~~~-.///             __.'              ~.   .~              `.__///           .'//                  \./                  \\`.///        .'//                     |                     \\`.///       .'// .-~"""""""~~~~-._     |     _,-~~~~"""""""~-. \\`.///     .'//.-"                 `-.  |  .-'                 "-.\\`.///   .'//______.============-..   \ | /   ..-============.______\\`./// .'______________________________\|/______________________________`.#pragma GCC optimize(2)#pragma comment(linker, "/STACK:102400000,102400000")#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=~0U>>1;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=1e5+10;const int maxx=1e6+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;//void readString(string &s)//{//static char str[maxn];//scanf("%s", str);//s = str;//}LL q[100][100];int a[100],cnt=0;LL F(LL a,LL i){LL l=1,r=a,b;W(l<r){LL mid=(l+r+1)/2;b=a;FOR(1,i,j) b/=mid;if(b) l=mid;else r=mid-1;}return l;}LL n;void solve(){S_1(n);if(n==1) {puts("-1");return ;}int i;for(i=60;i;i--) if(n>>i) break;FOR(1,i,j) if(n==1ll<<j) {puts("-1");return ;}FOR(1,i,j){LL root=F(n,j);LL xx=n;int tot1=0,tot2=0;while(xx%root==0)        {            xx/=root;            tot1++;        }        while(xx%(root+1)==0)        {            xx/=(root+1);            tot2++;        }        if(xx==1&&tot1+tot2==j)        {        a[++cnt]=j;    for(int k=1;k<=tot1;k++)                q[cnt][k]=root;            for(int k=tot1+1;k<=j;k++)                q[cnt][k]=root+1;        }    }    printf("%d\n",cnt);    for(int i=1;i<=cnt;i++)    {        printf("%d",a[i]);        for(int j=1;j<=a[i];j++)        {            printf(" %lld",q[i][j]);        }        puts("");    }}            int main(){    freopen( "little.in " , "r" , stdin );    freopen( "little.out" , "w" , stdout );    int t=1;    //init();    //s_1(t);    for(int cas=1;cas<=t;cas++)    {        //printf("Case #%d: ",cas);        solve();    }}