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

K  — DNA

Time Limit: 1 sec
Memory Limit: 32 MB

Alan was comleting his internship at chemistry faculty. At the beginning internship seemed to be easy. However one day dean of the faculty of chemistry have given John a certain task to solve. Alan was asked to perform DNA and protein modelling using free OpenEye software kit.

As Alan like all programmers is very lazy, he would like to use computer as much as possible. So now he needs to generate all possible DNA mutations and share them to OpenEye software. Alan decided to use computer not only for OpenEye software but also for mutation generation purpose.

At the beginning of the internship Alan studying a little bit of chemistry science, so he knows that DNA consisting of 4 elements (Adenine, Guanine, Cytosine, and Thymine) and can be described as a sequence of four letters (for example: GATCC). The K-th mutation of inital DNA sequence of length N is called a sequence that can be produced by replacing ((possibly to the same nucleotide)) exactly K elements of the sequence (for example GAT is the 1-st mutation of GGT and the 2-nd mutation of TTT).

Alan is given initial DNA sequence and maximal power of its possible mutataion. Can you produce all possible mutated DNA sequences for Alan?

INPUT

The number of tests T (T ≤ 50) is given on the first line. Each test case if described by two lines: one the first is number N (N ≤ 10 ) and K (K ≤ 5), on the second line is written DNA sequence of length N.

OUTPUT

For each test case print M — the number of different DNA mutations. After this print all mutated DNA sequences in alphabetical order. Refer to the sample output for details.

SAMPLE INPUT

1
3 1
AAA

SAMPLE OUTPUT

10
AAA
AAC
AAG
AAT
ACA
AGA
ATA
CAA
GAA
TAA

---

Problem by: Aleksej Viktorchik; Leonid Sislo
Huge Easy Contest #2

窮舉突變的位置，然後突變個數不可以大於 K，文章最後的 突變單字拼錯。

#include <stdio.h>
char s[20], buf[20];
int t, n, m, ans;
void dfs(int idx, int k) {
    if(k > m)   return;
    if(idx == n) {
        ans++;
        return;
    }
    buf[idx] = 'A';
    dfs(idx+1, k+(buf[idx] != s[idx]));
    buf[idx] = 'C';
    dfs(idx+1, k+(buf[idx] != s[idx]));
    buf[idx] = 'G';
    dfs(idx+1, k+(buf[idx] != s[idx]));
    buf[idx] = 'T';
    dfs(idx+1, k+(buf[idx] != s[idx]));
}
void dfs2(int idx, int k) {
    if(k > m)   return;
    if(idx == n) {
        buf[idx] = '\0';
        puts(buf);
        return;
    }
    buf[idx] = 'A';
    dfs2(idx+1, k+(buf[idx] != s[idx]));
    buf[idx] = 'C';
    dfs2(idx+1, k+(buf[idx] != s[idx]));
    buf[idx] = 'G';
    dfs2(idx+1, k+(buf[idx] != s[idx]));
    buf[idx] = 'T';
    dfs2(idx+1, k+(buf[idx] != s[idx]));
}
int main() {
    scanf("%d", &t);
    while(t--) {
        scanf("%d %d", &n, &m);
        scanf("%s", s);
        ans = 0;
        dfs(0, 0);
        printf("%d\n", ans);
        dfs2(0, 0);
    }
    return 0;
}