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

Integers 1, 2, 3,..., n are placed on a circle in the increasing order as in the following figure. We want to construct a sequence from these numbers on a circle. Starting with the number 1, we continually go round by picking out each k-th number and send to a sequence queue until all numbers on the circle are exhausted. This linearly arranged numbers in the queue are called Jump(n, k) sequence where 1n, k.

Let us compute Jump(10, 2) sequence. The first 5 picked numbers are 2, 4, 6, 8, 10 as shown in the following figure. And 3, 7, 1, 9 and 5 will follow. So we get Jump(10, 2) = [2,4,6,8,10,3,7,1,9,5]. In a similar way, we can get easily Jump(13, 3) = [3,6,9,12,2,7,11,4,10,5,1,8,13], Jump(13, 10) = [10,7,5,4,6,9,13,8,3,12,1,11,2] and Jump(10, 19) = [9,10,3,8,1,6,4,5,7,2].

You write a program to print out the last three numbers of Jump(n, k) for n, k given. For example suppose that n = 10, k = 2, then you should print 1, 9 and 5 on the output file. Note that Jump(1, k) = [1].

Input 

Your program is to read the input from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing two integers n and k, where 5n500, 000 and 2k500, 000.

Output 

Your program is to write to standard output. Print the last three numbers of Jump(n, k) in the order of the last third, second and the last first. The following shows sample input and output for three test cases.

Sample Input 

3 
10 2 
13 10 
30000 54321

Sample Output 

1 9 5 
1 11 2 
10775 17638 23432

---

為了加快模擬速度, 採用 zkw式ST, 單一測資是 O(nlogn)


#include <stdio.h>
int st[1<<20];
int main() {
    int n, k, i, j;
    int t;
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &k);
        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 = (k+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;
            if(i > n-3) {
                if(i > n-2) putchar(' ');
                printf("%d", last);
            }
            while(s) {
                st[s] --;
                s >>= 1;
            }
        }
        puts("");
    }
    return 0;
}