問題:
http://poj.org/problem?id=2005思路:
簡單題，但是容易錯
特殊情況: 兩張牌都是Ace，這也是Ace需要被當(dāng)作是1的唯一情況
代碼:
1 #include<stdio.h>
2 #include<stdlib.h>
3 #include<string.h>
4 int player, dealer;
5 int n, count[14];
6 char card[3][2];
7
8 int
9 score(char ch)
10 {
11 if(ch == 'A')
12 return 11;
13 else if(ch == 'T' || ch=='J' || ch=='Q' || ch=='K')
14 return 10;
15 else
16 return ch-'0';
17 }
18
19 int
20 main(int argc, char **argv)
21 {
22 int i, tmp, total;
23 while(scanf("%d", &n)!=EOF && n) {
24 total = player = dealer = 0; memset(count, 0, sizeof(count));
25 for(i=0; i<3; i++) {
26 scanf("%s", card[i]);
27 tmp = score(card[i][0]);
28 ++count[tmp];
29 if(i == 0)
30 dealer += tmp;
31 else
32 player += tmp;
33 }
34 if(player == 22) /* two Ace */
35 player = 12;
36 for(i=2; i<=11; i++) {
37 if((dealer+i==22 ? 12 : dealer+i) < player) {
38 if(i == 10)
39 total += (4*4*n-count[i]);
40 else
41 total += (4*n-count[i]);
42 }
43 }
44 printf("%.3f%%\n\n", ((double)total)/(52*n-3)*100);
45 }
46 }