hdu-5505 (number theory)

Topic Links:

GT and Numbers

Time limit:2000/1000 MS (java/others)

Memory limit:65536/65536 K (java/others)

problem DescriptionYou are given and numbersNandM.

Every step you can get a newNIn the to that multiplyNby a factor ofN.

Work out how many steps canN is equal to M at least.

If N can ' t is to M forever,print −1.

InputIn the first line there is a numberT.Tis the test number.

In the nextTLines there is numbersNandM.

T≤ ,1≤N≤1000000 ,1≤M≤2.

Be careful to the range of M.

You ' d better print the ' Enter ' and ' last line ' hack others.

You ' d better not print space in the last of all line when you hack others.

OutputFor each test case,output an answer.

Sample Input 31 11 22 4

Sample Output 0-11 Test Instructions:Ask a number n, each time multiply its factor, ask at least how many times can become m; Ideas:The n,m quality factor decomposition, if the M can be obtained from N, then the decomposition of M of the mass factor and N is the same as the mass factor, and the number of each factorization is not less than n; otherwise it cannot be obtained by n ;the minimum number of steps is of course for each factorization, greedy to get the corresponding mass in M number, the final answer is that all the number of factorization into the number of m requirements of the largest one; already mentally retarded to not write scanf; AC Code:

//#include <bits/stdc++.h>#include<iostream>#include<queue>#include<cmath>#include<map>#include<cstring>#include<algorithm>#include<cstdio>using namespacestd;#defineRiep (n) for (int i=1;i<=n;i++)#defineRIOP (n) for (int i=0;i<n;i++)#defineRJEP (n) for (int j=1;j<=n;j++)#defineRJOP (n) for (int j=0;j<n;j++)#defineMST (SS,B) memset (ss,b,sizeof (ss));typedef unsignedLong LongLL;ConstLL mod=1e9+7;Const DoublePi=acos (-1.0);Const intinf=0x3f3f3f3f;Const intn=1e6+8; template<classT>voidRead (t&num) {    CharCH;BOOLf=false;  for(Ch=getchar (); ch<'0'|| Ch>'9'; f= ch=='-', ch=GetChar ());  for(num=0; ch>='0'&&ch<='9'; num=num*Ten+ch-'0', ch=GetChar ()); F&& (num=-num);}intstk[ -], Tp;template<classT> Inlinevoidprint (T p) {if(!p) {Puts ("0");return; }  while(p) stk[++ TP] = p%Ten, p/=Ten;  while(TP) Putchar (stk[tp--] +'0'); Putchar ('\ n');}intPrime[n];structnode{inta[9],m;} Po[n];voidInit () { for(intI=1; i<n;i++) {po[i].m=0; }     for(intI=2; i<n;i++)    {        if(!Prime[i]) {PO[I].A[PO[I].M++]=i;  for(intj=2*i;j<n;j+=i) {po[j].a[po[j].m++]=i; PRIME[J]=1; }}}}ll Check (intx) {LL s=1;  for(intI=1; i<=x;i++) {s*=2; }    returns;}intGetans (intXinty) {    intL=0, R= (int) (Log (y/x+1)/log (2.0))+1;  while(l<=r) {intMid= (l+r) >>1; if(Check (mid) *x>=y) r=mid-1; ElseL=mid+1; }    returnr+1;}voidSolveintn,ll m) {            if(n==1)            {                if(m==1) printf ("0\n"); Elseprintf"-1\n"); }            Else            {                intans=0; inttempn=N; LL TEMPM=m;  for(intI=0; i<po[n].m;i++)                {                    intnum1=0;  while(tempn%po[n].a[i]==0) {TEMPN/=Po[n].a[i]; NUM1++; }                    intNum2=0;  while(tempm%po[n].a[i]==0) {tempm/=Po[n].a[i]; Num2++; }                    if(num2<NUM1) {printf ("-1\n"); return ; } ans=Max (Ans,getans (num1,num2)); }                if(tempm>1) printf ("-1\n"); Elseprintf"%d\n", ans); }}intMain () {Init (); intT; scanf ("%d",&T); intN;        LL m;  while(t--) {read (n); scanf ("%i64u",&m);        Solve (n,m); }    return 0;}

hdu-5505 (number theory)