Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

 

The Problem

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 every mth is 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 many 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 evil 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 to the ith prime.
 

The Input

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

The Output

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

Sample Input

6
0

Sample Output

4

---

拿線段樹來殺人, 做法 O (nlogn)

#include <stdio.h>

int Pt = 1, P[5500];
int sieve() {
    char mark[50000] = {};
    int i, j;
    for(i = 2; i < 50000; i++) {
        if(!mark[i]) {
            P[Pt++] = i;
            for(j = 2*i; j < 50000; j += i)
                mark[j] = 1;
        }
    }
}
int main() {
    sieve();
    int n, i, j, st[8193];
    while(scanf("%d", &n) == 1 && n) {
        int M;
        for(M = 1; M < n+1; M <<= 1);
        for(i = 2*M-1; i > 0; i--) {
            if(i >= M)
                st[i] = 1;
            else
                st[i] = st[i<<1]+st[i<<1|1];
        }
        int m, last, prev = 0, s;
        for(i = 1; i <= n; i++) {
            m = (P[i]+prev)%(n-i+1);
            if(m == 0)
                m = n-i+1;
            prev = m-1;
            for(s = 1; s < M;) {
                if(st[s<<1] < m)
                    m -= st[s<<1], s = s<<1|1;
                else
                    s = s<<1;
            }
            last = s-M+1;
            while(s) {
                st[s] --;
                s >>= 1;
            }
        }
        printf("%d\n", last);
    }
    return 0;
}