Hoj 1016 Joseph's problem I tree Array

Hoj 1016 Joseph's problem I

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	Source: Unknown
	Time Limit: 10 sec		Memory limit: 32 m

Submitted1484,Accepted439

The Joseph's problem is notoriously known. For those who are not familiar with the problem, among N people numbered 1, 2... n, standing in circle everyMTHIs going to be executed and only the life of the last remaining person will be saved. joseph was smart enough to choose the position of the last remaining person, thus saving his life to give the message about the incident.

Although good programmers have been saved since Joseph spread out this information, Joseph's cousin introduced a new variant of the malignant game. this insane character is known for its barbarian ideas and wishes to clean up the world from silly programmers. we had to infiltrate some the agents of the ACM in order to know the process in this new mortal game.

In order to save yourself from this edevil practice, you must develop a tool capable of predicting which person will be saved.

The destructive process

The persons are eliminated in a very peculiar order; M is a dynamical variable, which each time takes a different value corresponding to the prime numbers 'succession (2, 3, 5, 7 ...). so in order to kill the ith person, Joseph's cousin counts up toIthPrime.

Input

It consists of separate lines containing N [1 .. 3501], and finishes with a 0.

Output

The output will consist in separate lines containing the position of the person which life will be saved.

Sample Input

 
6
 

Sample output

 
4
 

Question:
Joseph's ring problem, but the interval between the K times is the K prime number.

Analysis:
Use the sieve method to filter out all prime numbers, and then use the tree array to calculate the K size. However, the data volume is large
If every input data is retrieved, it will be tle... So we can first store all the 3501
In an array, each query is directly output at O (1) time.

# Include <vector> # include <cstdio> # include <cstring> # include <iostream> using namespace STD; typedef long ll; # define debug puts ("here "); const int x = 3605; const int maxm = 36005; vector <int> prime; bool use [maxm]; int C [X]; void Init () {memset (use, false, sizeof (use); Prime. push_back (2); For (INT I = 3; I <maxm; I ++ = 2) if (! Use [I]) {Prime. push_back (I); For (Int J = I + I; j <maxm; j + = I) use [J] = true ;}} int lowbit (INT X) {return X &-X;} void modify (int x, int num) {While (x <X) {C [x] + = num; X + = lowbit (x) ;}} int query (int x) {int ret = 0; while (x> 0) {RET + = C [x]; x-= lowbit (x);} return ret;} void binary (int n, int K) {int L = 1, R = N; while (L <= r) {int mid = (L + r)> 1; int temp = query (MID); If (temp> = k) r = mid-1; else l = Mid + 1;} modify (L,-1);} int ans [X]; int main () {# ifndef online_judge freopen ("sum. in "," r ", stdin); // freopen (" sum. out "," W ", stdout); # endif int op = 3600; // controls Init (); int N; For (n = 1; n <op; N ++) {memset (C, 0, sizeof (c); For (INT I = 1; I <= N; I ++) Modify (I, 1 ); int pre = 1; int res = N; For (INT I = 0; I <n-1; I ++) {pre = (prime [I] + pre-2) % res + 1; binary (n, pre); res --;} For (INT I = 1; I <= N; I ++) if (C [I]) {ans [N] = I; break;} while (scanf ("% d", & N), n) printf ("% d \ n ", ans [N]); Return 0 ;}