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

Joseph likes taking part in programming contests. His favorite problem is, of course, Joseph's problem. It is stated as follows.

There are n persons numbered from 0 to n - 1 standing in a circle. The person number k, counting from the person number 0, is executed. After that the person number k of the remaining persons is executed, counting from the person after the last executed one. The process continues until only one person is left. This person is a survivor. The problem is, given n and k detect the survivor's number in the original circle.

Of course, all of you know the way to solve this problem. The solution is very short, all you need is one cycle:

     r := 0; 
     for i from 1 to n do 
         r := (r + k) mod i; 
     return r;

Here ``x mod y" is the remainder of the division of x by y

But Joseph is not very smart. He learned the algorithm, but did not learn the reasoning behind it. Thus he has forgotten the details of the algorithm and remembers the solution just approximately.

He told his friend Andrew about the problem, but claimed that the solution can be found using the following algorithm:

     r := 0; 
     for i from 1 to n do 
         r := r + (k mod i); 
     return r;

Of course, Andrew pointed out that Joseph was wrong. But calculating the function Joseph described is also very interesting.

Given n and k , find (k mod i) .

Input 

The input file contains several test cases, each of them consists of a line containing n and k (1n, k10⁹)

Output 

For each test case, output a line containing the sum requested.

Sample Input 

5 3

Sample Input 

7

---

try this code：
        for(int i = 1; i <= n; i++) {
            printf("%d : %d %d\n", i, k/i, k%i);
        }
n = 50, k = 50
1 : 50 0
2 : 25 0
3 : 16 2
4 : 12 2
5 : 10 0
6 : 8 2
7 : 7 1
8 : 6 2
9 : 5 5
10 : 5 0
11 : 4 6
12 : 4 2
13 : 3 11
14 : 3 8
15 : 3 5
16 : 3 2
17 : 2 16
18 : 2 14
19 : 2 12
20 : 2 10
21 : 2 8
22 : 2 6
23 : 2 4
24 : 2 2
25 : 2 0
26 : 1 24
27 : 1 23
28 : 1 22
29 : 1 21
30 : 1 20
31 : 1 19
32 : 1 18
33 : 1 17
34 : 1 16
35 : 1 15
36 : 1 14
37 : 1 13
38 : 1 12
39 : 1 11
40 : 1 10
41 : 1 9
42 : 1 8
43 : 1 7
44 : 1 6
45 : 1 5
46 : 1 4
47 : 1 3
48 : 1 2
49 : 1 1
50 : 1 0

我們發現，當 k/i 相同時，呈現一個 "等差級數"。
如此判定下 ... 計算之

#include <stdio.h>

int main() {
    long long n, k;
    while(scanf("%lld %lld", &n, &k) == 2) {
        long long l = 1, r, div;
        long long ret = 0;
        while(l <= k) {
            div = k/l;
            r = k/div;
            // d = div
            if(r > n)   r = n;
            long long a0 = k%l, d = -div, tn = r-l+1;
            ret += (a0*2+(tn-1)*d)*tn/2;
            //printf("%lld %lld %lld\n", l, r, k%l);
            l = r+1;
            if(r == n)
                break;
        }
        if(l <= n)
            ret += (n-l+1)*k;
        printf("%lld\n", ret);
    }
    return 0;
}
/*
999999999 999999999
999999999 999999998
999999990 999999999
*/