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PKU 2408 Anagram Groups (排序)

simplyzhao — Fri, 05 Nov 2010 07:38:00 GMT

Anagram Groups

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2318		Accepted: 649

Description

World-renowned Prof. A. N. Agram's current research deals with large anagram groups. He has just found a new application for his theory on the distribution of characters in English language texts. Given such a text, you are to find the largest anagram groups.

A text is a sequence of words. A word w is an anagram of a word v if and only if there is some permutation p of character positions that takes w to v. Then, w and v are in the same anagram group. The size of an anagram group is the number of words in that group. Find the 5 largest anagram groups.

Input

The input contains words composed of lowercase alphabetic characters, separated by whitespace(or new line). It is terminated by EOF. You can assume there will be no more than 30000 words.

Output

Output the 5 largest anagram groups. If there are less than 5 groups, output them all. Sort the groups by decreasing size. Break ties lexicographically by the lexicographical smallest element. For each group output, print its size and its member words. Sort the member words lexicographically and print equal words only once.

Sample Input

undisplayed
trace
tea
singleton
eta
eat
displayed
crate
cater
carte
caret
beta
beat
bate
ate
abet

Sample Output

Group of size 5: caret carte cater crate trace .
Group of size 4: abet bate beat beta .
Group of size 4: ate eat eta tea .
Group of size 1: displayed .
Group of size 1: singleton .

Source

Ulm Local 2000

思路:
這題將排序發(fā)揮到了極致啊呵呵，排序來(lái)排序去就AC了

代碼:

1 /* 47MS */
2 #include<stdio.h>
3 #include<stdlib.h>
4 #include<string.h>
5 #define MAX_NUM 30001
6 #define MAX_LEN 36
7 #define MAX_OUT 5
8 struct Word {
9     char word[MAX_LEN];
10     char word_cmp[MAX_LEN];
11 } words[MAX_NUM];
12
13 struct Summary {
14     struct Word *first;
15     int count;
16 } smmry[MAX_NUM];
17
18 int total, total_category;
19
20 int
21 cmp_char(const void *arg1, const void *arg2)
22 {
23     return (*(char *)arg1) - (*(char *)arg2);
24 }
25
26 int
27 cmp_words(const void *arg1, const void *arg2)
28 {
29     int ret = strcmp(((struct Word *)arg1)->word_cmp, ((struct Word *)arg2)->word_cmp);
30     if(ret == 0)
31         ret = strcmp(((struct Word *)arg1)->word, ((struct Word *)arg2)->word);
32     return ret;
33 }
34
35 int
36 cmp_category(const void *arg1, const void *arg2)
37 {
38     int ret = ((struct Summary *)arg2)->count - ((struct Summary *)arg1)->count;
39     if(ret == 0)
40         ret = strcmp(((struct Summary *)arg1)->first->word, ((struct Summary *)arg2)->first->word);
41     return ret;
42 }
43
44 int
45 main(int argc, char **argv)
46 {
47     int i, j, num, len;
48     total = total_category = 0;
49     while(scanf("%s", words[total].word) != EOF) {
50         len = strlen(words[total].word);
51         strcpy(words[total].word_cmp, words[total].word);
52         qsort(words[total].word_cmp, len, sizeof(char), cmp_char);
53         ++total;
54     }
55     qsort(words, total, sizeof(struct Word), cmp_words);
56
57     num = 1;
58     for(i=1; i<total; i++) {
59         if(strcmp(words[i].word_cmp, words[i-1].word_cmp) == 0)
60             ++num;
61         else {
62             smmry[total_category].first = words+i-num;
63             smmry[total_category].count = num;
64             ++total_category;
65             num = 1;
66         }
67     }
68     smmry[total_category].first = words+i-num;
69     smmry[total_category++].count = num;
70     qsort(smmry, total_category, sizeof(struct Summary), cmp_category);
71
72     total_category = total_category < MAX_OUT ? total_category : MAX_OUT;
73     for(i=0; i<total_category; i++) {
74         printf("Group of size %d: %s ", smmry[i].count, smmry[i].first->word);
75         for(j=1; j<smmry[i].count; j++)
76             if(strcmp((smmry[i].first+j)->word, (smmry[i].first+j-1)->word) != 0)
77                 printf("%s ", (smmry[i].first+j)->word);
78         printf(".\n");
79     }
80 }

simplyzhao 2010-11-05 15:38 發(fā)表評(píng)論

PKU 2418 Hardwood Species

simplyzhao — Fri, 29 Oct 2010 16:59:00 GMT

問(wèn)題:
http://poj.org/problem?id=2418

思路:
題意清晰明了，不難，用三種方法分別實(shí)現(xiàn):
快速排序
動(dòng)態(tài)生成節(jié)點(diǎn)的二叉查找樹
靜態(tài)分配節(jié)點(diǎn)的二叉查找樹

結(jié)果發(fā)現(xiàn)，原來(lái)對(duì)于快速排序與靜態(tài)分配節(jié)點(diǎn)都不是很熟悉，二維數(shù)組的快速排序分析見http://m.shnenglu.com/Joe/archive/2010/10/29/131746.html，而動(dòng)態(tài)生成節(jié)點(diǎn)則需要注意如果函數(shù)需要修改指針，那么必須傳遞指向指針的指針，因?yàn)镃是值傳遞的

另外，我以為靜態(tài)分配節(jié)點(diǎn)應(yīng)該比動(dòng)態(tài)分配節(jié)點(diǎn)節(jié)約很多時(shí)間的，結(jié)果居然差不多...而快速排序在這題明顯是最耗時(shí)的

代碼(快排)

1 /* 35640K 1844MS */
2 #include<stdio.h>
3 #include<stdlib.h>
4 #include<string.h>
5 #define MAX_LEN 36
6 #define MAX_NUM 1000001
7 char species[MAX_NUM][MAX_LEN];
8
9 int
10 cmp(const void *arg1, const void *arg2)
11 {
12     return strcmp((char *)arg1, (char *)arg2);
13 }
14
15 int
16 main(int argc, char **argv)
17 {
18     int i, count, total = 0;
19     while(gets(species[total]) != NULL)
20         ++total;
21     qsort(species, total, sizeof(species[0]), cmp);
22     count = 1;
23     for(i=1; i<total; i++) {
24         if(strcmp(species[i], species[i-1]) == 0)
25             ++count;
26         else {
27             printf("%s %.4f\n", species[i-1], (count*100.0)/total);
28             count = 1;
29         }
30     }
31     printf("%s %.4f\n", species[total-1], (count*100.0)/total);
32 }

代碼(動(dòng)態(tài)分配節(jié)點(diǎn)的BST，原本想實(shí)現(xiàn)下destroy函數(shù)的，結(jié)果怕麻煩就留給系統(tǒng)回收吧(*^__^*) 嘻嘻……)

1 /* binary search tree(dynamic allocation) */
2 /* 544K 1188MS */
3 #include<stdio.h>
4 #include<stdlib.h>
5 #include<string.h>
6 #include<assert.h>
7 #define MAX_LEN 36
8 struct Node {
9     char spec[MAX_LEN];
10     int count;
11     struct Node *left, *right;
12 };
13 int total;
14
15 struct Node *
16 create_node(char *str)
17 {
18     struct Node *node = (struct Node *)malloc(sizeof(struct Node));
19     assert(node != NULL);
20     strcpy(node->spec, str);
21     node->left = node->right = NULL;
22     node->count = 1;
23     return node;
24 }
25
26 void
27 insert(struct Node **root, char *str)
28 {
29     int ret;
30     struct Node *node;
31     if(*root==NULL) {
32         *root = create_node(str);
33         return;
34     }
35     ret = strcmp((*root)->spec, str);
36     if(ret == 0)
37         ++((*root)->count);
38     else if(ret < 0)
39         insert(&((*root)->right), str);
40     else
41         insert(&((*root)->left), str);
42 }
43
44 void
45 inorder(struct Node *root)
46 {
47     if(root == NULL)
48         return;
49     inorder(root->left);
50     printf("%s %.4f\n", root->spec, (root->count)*100.0/total);
51     inorder(root->right);
52 }
53
54 void
55 destroy(struct Node **root)
56 {
57 }
58
59 int
60 main(int argc, char **argv)
61 {
62     char str[MAX_LEN];
63     struct Node *bst = NULL;
64     total = 0;
65     while(gets(str) != NULL) {
66         ++total;
67         insert(&bst, str);
68     }
69     inorder(bst);
70 }

代碼(靜態(tài)分配節(jié)點(diǎn)的BST)

1 /* binary search tree(static allocation) */
2 /* 492K 1188MS */
3 #include<stdio.h>
4 #include<stdlib.h>
5 #include<string.h>
6 #include<assert.h>
7 #define MAX_LEN 36
8 #define MAX_NUM 10007
9 #define ROOT 1
10 struct Node {
11     char spec[MAX_LEN];
12     int count;
13     int left, right;
14 }bst[MAX_NUM];
15 int cur_index, total;
16
17 int
18 find(int root, char *str)
19 {
20     int ret, parent, cur = root;
21     while(cur != 0) {
22         parent = cur;
23         ret = strcmp(bst[cur].spec, str);
24         if(ret == 0) {
25             ++bst[cur].count;
26             return 0;
27         } else if(ret < 0) {
28             cur = bst[cur].right;
29         } else {
30             cur = bst[cur].left;
31         }
32     }
33     return parent;
34 }
35
36 #define ADD(index, str) { \
37     strcpy(bst[index].spec, str); \
38     bst[index].left = bst[index].right = 0; \
39     bst[index].count = 1; \
40     ++index; }
41
42 void
43 insert(int parent, char *str)
44 {
45     int ret = strcmp(bst[parent].spec, str);
46     assert(ret != 0);
47     if(ret < 0)
48         bst[parent].right = cur_index;
49     else
50         bst[parent].left = cur_index;
51     ADD(cur_index, str);
52 }
53
54 void
55 inorder(int index)
56 {
57     if(index == 0)
58         return;
59     inorder(bst[index].left);
60     printf("%s %.4f\n", bst[index].spec, (bst[index].count*100.0)/total);
61     inorder(bst[index].right);
62 }
63
64 int
65 main(int argc, char **argv)
66 {
67     int parent;
68     char str[MAX_LEN];
69     total = 1;
70     cur_index = ROOT;
71     gets(str);
72     ADD(cur_index, str); /* create the root node first */
73     while(gets(str) != NULL) {
74         ++total;
75         if((parent=find(ROOT, str)) > 0)
76             insert(parent, str);
77     }
78     inorder(ROOT);
79 }

simplyzhao 2010-10-30 00:59 發(fā)表評(píng)論

[Tips][Original] qsort應(yīng)用于指針數(shù)組與二維數(shù)組(字符)的差異

simplyzhao — Fri, 29 Oct 2010 07:09:00 GMT

在將qsort函數(shù)應(yīng)用于對(duì)指針數(shù)組與二維數(shù)組排序時(shí)，傳遞給compare函數(shù)的參數(shù)類型是不同的

首先，我們舉個(gè)簡(jiǎn)單的例子，先將qsort對(duì)整數(shù)數(shù)組排序:

1 int
2 cmp(const void *arg1, const void *arg2)
3 {
4     return (*(int *)arg1)-(*(int *)arg2);
5 }
6
7 int
8 main(int argc, char **argv)
9 {
10     int i;
11     int arr[] = {3, 1, 5, 2, 4};
12     qsort(arr, sizeof(arr)/sizeof(arr[0]), sizeof(int), cmp);
13 }

排序針對(duì)的是數(shù)組里的元素而言的，這里整數(shù)數(shù)組的元素就是整數(shù)，因此qsort的第三個(gè)參數(shù)就是sizeof(int)，而傳遞給比較函數(shù)cmp的參數(shù)就是相對(duì)應(yīng)的指向整數(shù)的指針

接著，我們來(lái)看看指針數(shù)組的情形:

1 int
2 cmp(const void *arg1, const void *arg2)
3 {
4     return strcmp((*(char **)arg1), (*(char **)arg2));
5 }
6
7 int
8 main(int argc, char **argv)
9 {
10     int i;
11     /* pointer array */
12     char *str[] = {"java", "c", "python", "perl"};
13     qsort(str, sizeof(str)/sizeof(str[0]), sizeof(char *), cmp);
14 }

這里的理解其實(shí)跟整數(shù)數(shù)組差不多，關(guān)鍵是抓住數(shù)組里元素的類型，既然稱之為指針數(shù)組，那數(shù)組元素的類型自然就是指針，因此qsort的第三個(gè)參數(shù)就是sizeof(char *)，而傳遞給比較函數(shù)cmp的參數(shù)就是相對(duì)應(yīng)的指向指針的指針，這里即char **類型

二維數(shù)組的理解最為復(fù)雜，代碼如下:

1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4
5 int
6 cmp1(const void *arg1, const void *arg2)
7 {
8     return strcmp((*((char (*)[])arg1)), (*((char (*)[])arg2)));
9 }
10
11 int
12 cmp2(const void *arg1, const void *arg2)
13 {
14     return strcmp((char *)arg1, (char *)arg2);
15 }
16
17 int
18 main(int argc, char **argv)
19 {
20     int i;
21     char str1[4][8] = {"java", "c", "python", "peal"};
22     printf("COMPARE-FUNCTION-1\n");
23     qsort(str1, 4, sizeof(str1[0]), cmp1);
26
27     char str2[4][8] = {"java", "c", "python", "peal"};
28     printf("COMPARE-FUNCTION-2\n");
29     qsort(str2, 4, sizeof(str2[0]), cmp2);
34 }

這里cmp1與cmp2都能正常的工作(*^__^*) 嘻嘻……
還是按照上述方法來(lái)分析，抓住數(shù)組元素的類型來(lái)入手，二維數(shù)組實(shí)際上就是數(shù)組的數(shù)組，因此二維數(shù)組的元素類型就是一維數(shù)組，因此qsort的第三個(gè)參數(shù)就是sizeof(str1[0])或sizeof(str2[0])，那傳遞給比較函數(shù)的參數(shù)應(yīng)該就是指向數(shù)組的指針，這點(diǎn)可以通過(guò)gdb設(shè)置斷點(diǎn)來(lái)得到證實(shí):

1 (gdb) p &str1[0]
2 $1 = (char (*)[8]) 0xbffff2cc
3 (gdb) p &str1[1]
4 $2 = (char (*)[8]) 0xbffff2d4
5
6 Breakpoint 2, cmp1 (arg1=0xbffff2cc, arg2=0xbffff2d4) at char_test2.c:8
7 8 return strcmp((*((char (*)[])arg1)), (*((char (*)[])arg2)));
8 (gdb) p arg1
9 $3 = (const void *) 0xbffff2cc
10 (gdb) p arg2
11 $4 = (const void *) 0xbffff2d4
12 (gdb) p *(char (*)[])arg1
13 $5 = "j"
14 (gdb) p *(char (*)[8])arg1
15 $6 = "java\000\000\000"

通過(guò)第2行與第9行的比較可以發(fā)現(xiàn)，比較函數(shù)的參數(shù)arg1其實(shí)就是&str1[0]，類型為char (*)[]，這也是為什么cmp1能正常工作的原因
那么cmp2呢，它為什么正確呢?
在cmp1中：strcmp((*((char (*)[])arg1)), (*((char (*)[])arg2))); 這里傳遞給strcmp的參數(shù)之所以不會(huì)出錯(cuò)，是因?yàn)槲覀儗rg1解地址操作獲得一個(gè)數(shù)組，而數(shù)組名其實(shí)是指向數(shù)組首元素的指針，arg1既然是指向str1[0]這個(gè)一維數(shù)組的指針，而str1[0]本身其實(shí)就是指向這個(gè)一維數(shù)組的指針，也就是說(shuō)arg1其實(shí)就是str1[0]，因此cmp2能夠正常工作

1 (gdb) p &str1[0]
2 $3 = (char (*)[8]) 0xbffff2cc
3 (gdb) p &str1[0][0]
4 $4 = 0xbffff2cc "java"
5 (gdb) p arg1
6 $5 = (const void *) 0xbffff2cc
7 (gdb) p (char *)arg1
8 $6 = 0xbffff2cc "java"

額...貌似越說(shuō)越復(fù)雜的樣子，不過(guò)這是我理解的過(guò)程，見諒...

simplyzhao 2010-10-29 15:09 發(fā)表評(píng)論

PKU 1089 Intervals

simplyzhao — Sun, 17 Oct 2010 11:35:00 GMT

問(wèn)題:
http://poj.org/problem?id=1089

思路:
根據(jù)interval的begin升序排列，然后對(duì)于interval A, B只存在三種情況:
A包含B、A與B相交、A與B分離

代碼:

1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 #define MAX_LEN 50001
5 struct Interval {
6     int begin, end;
7 }itvs[MAX_LEN], target[MAX_LEN];
8 int N, total;
9
10 int
11 compare(const void *arg1, const void *arg2)
12 {
13     struct Interval *a = (struct Interval *)arg1;
14     struct Interval *b = (struct Interval *)arg2;
15     if(a->begin == b->begin)
16         return a->end - b->end;
17     return a->begin - b->begin;
18 }
19
20 void
21 init()
22 {
23     int i;
24     for(i=0; i<N; i++)
25         scanf("%d %d", &itvs[i].begin, &itvs[i].end);
26     qsort(itvs, N, sizeof(struct Interval), compare);
27     total = 0;
28 }
29
30 void
31 solve()
32 {
33     int i;
34     struct Interval cur = itvs[0];
35     for(i=1; i<N; i++) {
36         if(itvs[i].begin > cur.end) {
37             target[total++] = cur;
38             cur = itvs[i];
39         } else {
40             if(itvs[i].end > cur.end)
41                 cur.end = itvs[i].end;
42         }
43     }
44     target[total++] = cur;
45 }
46
47 void
48 output()
49 {
50     int i;
51     for(i=0; i<total; i++)
52         printf("%d %d\n", target[i].begin, target[i].end);
53 }
54
55 int
56 main(int argc, char **argv)
57 {
58     while(scanf("%d", &N) != EOF) {
59         init();
60         solve();
61         output();
62     }
63 }

simplyzhao 2010-10-17 19:35 發(fā)表評(píng)論

PKU 1828 Monkeys' Pride

simplyzhao — Wed, 08 Sep 2010 15:41:00 GMT

問(wèn)題:
http://acm.pku.edu.cn/JudgeOnline/problem?id=1828

思路:
按照坐標(biāo)從上到下、從左到右排序
首先想到的是O(n^2)的算法，時(shí)間需要1600+MS
然后，發(fā)現(xiàn)其實(shí)在排序之后只要從后向前掃描一遍即可得出結(jié)果

代碼:

1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 #define MAX_LEN 50001
5 int n;
6 struct Point {
7     int x, y;
8 }points[MAX_LEN];
9
10 int
11 compare(const void *arg1, const void *arg2)
12 {
13     struct Point *a = (struct Point *)arg1;
14     struct Point *b = (struct Point *)arg2;
15     if(a->x == b->x)
16         return a->y - b->y;
17     return a->x - b->x;
18 }
19
20 int
21 main(int argc, char **argv)
22 {
23     int i, j, cnt, ymax;
24     while(scanf("%d", &n)!=EOF && n) {
25         for(i=0; i<n; i++)
26             scanf("%d %d", &points[i].x, &points[i].y);
27         qsort(points, n, sizeof(struct Point), compare);
28         /* O(n^2) AC 1600+MS
29         cnt = 0;
30         for(i=0; i31             for(j=i+1; j32                 if(points[j].y >= points[i].y)
33                     break;
34             }
35             if(j == n)
36                 ++cnt;
37         }
38         */
39         /* O(nlgn) AC 235MS */
40         cnt = 1;
41         ymax = points[n-1].y;
42         for(i=n-2; i>=0; i--) {
43             if(ymax < points[i].y) {
44                 ++cnt;
45                 ymax = points[i].y;
46             }
47         }
48         printf("%d\n", cnt);
49     }
50 }

simplyzhao 2010-09-08 23:41 發(fā)表評(píng)論

PKU 2388 Who's in the Middle

simplyzhao — Sun, 04 Jul 2010 02:19:00 GMT

問(wèn)題:
http://acm.pku.edu.cn/JudgeOnline/problem?id=2388

思路:
1. 排序
這題直接用快排就能AC，很水...

2. 求解第i個(gè)順序統(tǒng)計(jì)量(詳見CLRS)
利用快速排序過(guò)程中的思路，期望時(shí)間復(fù)雜度是O(n)

ps: partition函數(shù)的執(zhí)行過(guò)程是比較有意思的呵呵

1 int
2 partition(long *arr, int begin, int end)
3 {
4     int i, j;
5     i = begin;
6     long pivot = arr[begin];
7     for(j=begin+1; j<=end; j++)
8         if(arr[j] <= pivot)
9             swap(arr, ++i, j);
10     swap(arr, i, begin);
11     return i;
12 }
13
14 long
15 find_ith_os(long *arr, int begin, int end, int ith)
16 {
17     if(begin == end)
18         return arr[begin];
19     int p = partition(arr, begin, end);
20     int k = p-begin+1;
21     if(k == ith)
22         return arr[p];
23     else if(ith < k)
24         return find_ith_os(arr, begin, p-1, ith);
25     else
26         return find_ith_os(arr, p+1, end, ith-k);
27 }

simplyzhao 2010-07-04 10:19 發(fā)表評(píng)論

PKU 1007 DNA Sorting/2299 Ultra-QuickSort

simplyzhao — Sun, 04 Jul 2010 02:07:00 GMT

問(wèn)題:
http://acm.pku.edu.cn/JudgeOnline/problem?id=1007
http://acm.pku.edu.cn/JudgeOnline/problem?id=2299

思路:
求逆序?qū)Φ膫€(gè)數(shù)
這兩題的基本問(wèn)題是一致的，給定一個(gè)數(shù)組(包括字符串)，求出逆序?qū)Φ膫€(gè)數(shù)

1. 最簡(jiǎn)單的方法
兩層循環(huán)，復(fù)雜度O(n^2)

1 int
2 inversion_cal(char *str)
3 {
4     int i, j, count = 0;
5     int len = strlen(str);
6     for(i=0; i<len; i++)
7         for(j=i+1; j<len; j++)
8             if(str[i] > str[j])
9                 ++count;
10     return count;
11 }

2. 歸并排序＆分治思想
其實(shí)，只要將歸并排序稍加修改，就是一個(gè)求解逆序?qū)€(gè)數(shù)問(wèn)題的O(nlgn)方法
要理解的是這其中涉及的分治思想(三步驟):
a. 分解為子問(wèn)題
b. 求解子問(wèn)題
c. 合并子問(wèn)題的解來(lái)得到原問(wèn)題的解
具體對(duì)應(yīng)到求逆序?qū)€(gè)數(shù)的問(wèn)題:
a. 將原數(shù)組分解為前后兩個(gè)子數(shù)組
b. 求解子數(shù)組的逆序?qū)€(gè)數(shù)
c. 合并前后子數(shù)組，同時(shí)計(jì)算逆序?qū)€(gè)數(shù)(這個(gè)是指逆序?qū)Φ牡谝粋€(gè)數(shù)在前子數(shù)組中，而第二個(gè)數(shù)在后子數(shù)組中)

1 long long
2 merge_count(long *arr, long *temp, long begin, long end)
3 {
4     if(begin >= end)
5         return 0;
6     long i, j, k, mid = (begin+end)/2;
7     long long rt = 0;
8     rt += merge_count(arr, temp, begin, mid);
9     rt += merge_count(arr, temp, mid+1, end);
10     i = k = begin;
11     j = mid+1;
12     while(i<=mid && j<=end) {
13         if(arr[i] <= arr[j])
14             temp[k++] = arr[i++];
15         else {
16             temp[k++] = arr[j++];
17             rt += (mid-i+1);
18         }
19     }
20     for( ; i<=mid; i++)
21         temp[k++] = arr[i];
22     for( ; j<=end; j++) {
23         temp[k++] = arr[j];
24         rt += (mid-i+1);
25     }
26     /* copy */
27     for(k=begin; k<=end; k++)
28         arr[k] = temp[k];
29     return rt;
30 }

3. 特例方法
針對(duì)PKU 1007該題的特殊性: 字符串中只包含A, G, T, C四個(gè)字母，還有一種更加簡(jiǎn)單的O(n)方法

1 int
2 inversion_cal2(char *str)
3 {
4     int i, temp[4], count = 0;
5     int len = strlen(str);
6     memset(temp, 0, sizeof(temp));
7     for(i=len-1; i>=0; i--) {
8         switch(str[i]) {
9             case 'A':
10                 ++temp[0];
11                 break;
12             case 'C':
13                 ++temp[1];
14                 count += temp[0];
15                 break;
16             case 'G':
17                 ++temp[2];
18                 count += (temp[0]+temp[1]);
19                 break;
20             case 'T':
21                 ++temp[3];
22                 count += (temp[0]+temp[1]+temp[2]);
23                 break;
24         }
25     }
26     return count;
27 }

simplyzhao 2010-07-04 10:07 發(fā)表評(píng)論