Description according to some books, God's one failure of the creation experience is this: the first day,God created the basic elements of a world, called "Yuan". The next day,God created a new element, called "alpha". "Alpha" is defined as a set of "meta" components. It is easy to see that there are two different "alpha" types. On the third day,God has created a new element called "beta". "Beta" is defined as a set of "alpha" components. It is easy to find a total of four different "beta". Day Fourth,God created the new element "gamma", "gamma" is defined as the set of "beta". Obviously, there will be 16 different "gamma" altogether. If this goes on, the fourth element God created will be 65536, and the fifth element will have 2^65536. This is going to be an astronomical one. However, God did not expect the growth of the number of elements to be so rapid. He wants to enrich the elements of the world, so, day after day, year after year, he repeats the creation of new elements ... But soon, when God created the last element, "Theta," he found that there were so many elements in the world that the world's capacity was inadequate and unbearable. So on this day, God destroys the world. So far, God remembers the failure of the creation experience, and now he wants to ask you how many of the elements he created last time, "Theta"? God thinks the number may be too large to be expressed, so you just need to answer the value of this number to P modulo. You can think of God from "alpha" to "theta" altogether creating 10^9 elements, or 10^18 times, or simply ∞ times. Sentence Test instructions:
Input Next T line, a positive integer p per line, represents the value you need to modulo
Outputt line, one positive integer per line, for the value after the answer p is modulo
Sample Input3
2
3
6Sample Output0
1
4HINT
For 100% of the data, T<=1000,p<=10^7 expands Euler's theorem: for any natural number a,n,p is established. Set Ans=f[p], because for any [1,10^7] integer, it can be up to a LOGN operation will become 1, so the violent demand for a single O (sqrt (n) logn).
#include <cstdio>#include<cctype>#include<queue>#include<cmath>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i;i=next[i])using namespaceStd;inlineintRead () {intx=0, f=1;CharC=GetChar (); for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1; for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;} typedefLong Longll;intPow (ll N,intMintp) {ll ans=1; while(m) m&1? (ans*=n)%=p:1, (n*=n)%=p,m>>=1; return(int) ans;intPhiintN) {intans=N; for(intI=2; i*i<=n;i++)if(n%i==0) { while(n%i==0) n/=i; Ans=ans/i* (I-1); } if(n>1) ans=ans/n* (n1); returnans;}intFintN) {if(n==1)return 0; intC=Phi (n); returnPow2, c+f (c), n);}intMain () {Dwn (T,read (),1) printf ("%d\n", F (read ())); return 0;}View Code
BZOJ3884: The correct use of God and set
