UVA 11353 - A Different Kind of Sorting Shipu Ahamed - 01 Sep 2020 Problem PDFSolution:/****************************************************************** *** Problem : *** *** Author : Shipu Ahamed (Psycho Timekiller) *** *** E-mail : shipuahamed01@gmail.com *** *** University : BUBT,Dept. of CSE *** *** Team : BUBT_Psycho *** *** My Blog : http://shipuahamed.blogspot.com *** *** Facebook : http://www.facebook.com/DeesheharaShipu *** ******************************************************************/ #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; #define pi 2*acos(0.0) #define all(v) v.begin(),v.end() //input #define si(t) scanf("%d",&t) #define sl(t) scanf("%lld",&t) #define sf(t) scanf("%f",&t) #define sd(t) scanf("%lf",&t) #define sc(c) scanf("%c",&c) #define sii(a,b) scanf("%d%d",&a,&b) #define sll(a,b) scanf("%lld%lld",&a,&b) //Output #define P(a) printf("%d\n",a) #define PL(a) printf("%lld\n",a) #define PF(a) printf("%f\n",a) #define PDB(a) printf("%lf\n",a) #define PN(a) printf("%d ",a) #define PLN(a) printf("%lld ",a) #define PFN(a) printf("%f ",a) #define PDBN(a) printf("%lf ",a) #define PP(a,b) printf("%d %d\n",a,b) #define PPN(a,b) printf("%d %d ",a,b) #define PPL(a,b) printf("%lld %lld\n",a,b) #define PPLN(a,b) printf("%lld %lld ",a,b) #define CP(a) cout< pii; typedef pair piii; typedef pair pll; typedef pair psi; typedef vector vi; typedef vector vll; typedef vector vpii; typedef vector vs; typedef set si; typedef set ss; typedef map mii; typedef map mll; typedef map msi; template string toString( T Number ){stringstream st;st << Number;return st.str();} template T SOD(T n) {__typeof(n) sum=0;for(__typeof(n) i=1;i*i<=n;i++)sum+=(n%i)?0:((i*i==n)?i:i+n/i);return sum;} //For Define #define forab(i,a,b) for(__typeof(b) i=(a);i<=(b);i++) #define for0(i,n) forab(i,0,(n)-1) #define for1(i,n) forab(i,1,n) #define rforba(i,b,a) for(__typeof(b) i=(b);i>=(a);i--) #define rfor0(i,n) rforba(i,(n)-1,0) #define rfor1(i,n) rforba(i,n,1) #define forstl(i,s) for(__typeof((s).end()) i=(s).begin(); i != (s).end(); i++) //Debug #define dbg(x) cout << #x << " -> " << (x) << endl; #define dbgsarr(i,a) cout<<#a<<"["< "< "< "<>p ; return p;} ll pow(ll a,ll b, ll m) { ll res = 1; while(b) { if(b & 1) { res = ( (res % m) * (a % m) ) %m ; } a= ((a%m) * (a%m)) %m; b >>= 1; } return res; } ll modInverse(ll a, ll m){return pow(a,m-2,m);} ////============ CONSTANT ===============//// #define mx7 10000007 #define mx6 2000006 #define mx5 100005 #define inf 1<<30 //infinity value #define eps 1e-9 #define mx (1500) #define mod 1000000007 ////=====================================//// struct factcount { ll value=0,cnt=0; }A[mx6]; bool prime[mx6+10]; ll pd[mx6+10]; int k=1; void seive() { for(int i=4;i<=mx6;i+=2) prime[i]=1; prime[0]=1; prime[1]=1; pd[0]=2; for(ll i=3;i<=mx6-6;i+=2) { if(!prime[i]) { pd[k++]=i; ll mul = i<<1; ll j=i; for(j*=i;j<=mx6;j+=mul) { prime[j]=1; } } } // PN(k),PL(pd[k-1]); } void fact(int n) { int p=n; A[p].value=n; int sq = sqrt(n),i=0; int c=0; while(n!=1) { if(pd[i]>sq || i==k-1) break; while(n%pd[i]==0) { n/=pd[i]; c++; } i++; } if(n!=1) { c++; } A[p].cnt=c; } bool cmp(factcount a, factcount b) { if(a.cnt==0 || b.cnt==0) return false; if(a.cnt==b.cnt) return a.valuehttps://github.com/Shipu/OnlineJudgeProblemSolutionWithCPlusPlus/tree/master/uva/11353/11353.cpp