poj训练计划-3(3295/2993/2996)

 1 #include <stdio.h>
 2 
 3 int a[5];
 4 char str[500];
 5 
 6 int type(char c) {
 7     if (c == ‘K‘ || c == ‘A‘ || c == ‘C‘ || c == ‘E‘) {
 8         return 2;
 9     }
10     if (c == ‘N‘) {
11         return 1;
12     }
13     return 0;
14 }
15 
16 int calculate(char *s, int &r) {
17     int q = type(s[0]);
18     int z1, z2;
19     int k1, k2;
20     if (q == 2) {
21         k1 = calculate(s + 1, z1);    
22         k2 = calculate(s + 1 + z1, z2);
23         r = z1 + z2 + 1;
24         switch (s[0]) {
25             case ‘K‘:
26                 return k1 && k2;
27             case ‘A‘:
28                 return k1 || k2;
29             case ‘C‘:
30                 return !k1 || k2;
31             case ‘E‘:
32                 return k1 == k2;
33         }
34     } else if (q == 1) {
35         k1 = calculate(s + 1, z1);
36         r = z1 + 1;
37         return 1 - k1;
38     }
39     r = 1;
40     switch (s[0]) {
41         case ‘p‘:
42             return a[0];
43         case ‘q‘:
44             return a[1];
45         case ‘r‘:
46             return a[2];
47         case ‘s‘:
48             return a[3];
49     }
50     return a[4];
51 }
52 
53 
54 int search(int w) {
55     if (w == 5) {    
56         int r;
57         if (calculate(str, r) != 1) {
58             return 0;
59         }
60         return 1;
61     }
62     for (int i = 0; i < 2; i++) {
63         a[w] = i;
64         if (!search(w + 1)) {
65             return 0;
66         }
67     }
68     return 1;
69 }
70 
71 int main() {
72     while (~scanf("%s", str) && str[0] != ‘0‘) {
73         printf("%s\n", search(0) ? "tautology" : "not");
74     }
75     return 0;
76 }

 1 #include <stdio.h>
 2 
 3 char s[100];
 4 char str[2][6] = {"White", "Black"};
 5 char ch[2][7] = {"KQRBNP", "kqrbnp"};
 6 int w[2][6];
 7 int ans[2][6][65][2];
 8 char f[9][9];
 9 
10 void print(int k) {
11     int flag = 0;
12     printf("%s: ", str[k]);
13     for (int i = 0; i < 6; i++) { 
14         for (int j = 0; j < w[k][i]; j++) { 
15             if (flag) {
16                 printf(",");
17             }
18             flag = 1;
19             if (i < 5) {
20                 printf("%c", ch[0][i]);
21             }
22             printf("%c%d", ans[k][i][j][1] + ‘a‘, ans[k][i][j][0] + 1);
23         }
24     }
25 }
26 
27 int type(char c) {
28     for (int i = 0; i < 2; i++) 
29         for (int j = 0; j < 6; j++) {
30             if (ch[i][j] == c) {
31                 return i * 10 + j;
32             }
33         }
34     return -1;
35 }
36 
37 int main() {
38     for (int i = 0; i < 17; i++) {
39         scanf("%s", s);
40         if (i % 2) { 
41             for (int j = 0; j < 8; j++) {    
42                 int z = j * 4 + 2;
43                 f[7 - i / 2][j] = s[z];    
44             }
45         }
46     }
47     for (int i = 0; i < 8; i++)
48         for (int j = 0; j < 8; j++) {
49             int t = type(f[i][j]);
50             if (t != -1 && t < 10) {
51                 ans[0][t][w[0][t]][0] = i;
52                 ans[0][t][w[0][t]++][1] = j;
53             }
54         }
55     for (int i = 7; i >= 0; i--)
56         for (int j = 0; j < 8; j++) {
57             int t = type(f[i][j]);
58             if (t > 9) {
59                 t -= 10;
60                 ans[1][t][w[1][t]][0] = i;
61                 ans[1][t][w[1][t]++][1] = j;
62             }
63         }
64     print(0);
65     printf("\n");
66     print(1);
67     printf("\n");
68 }

 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 char ch[2][7] = {"KQRBNP", "kqrbnp"};
 5 char s[20][50];
 6 char r[1000];
 7 
 8 int type(char c) {
 9     for (int i = 0; i < 6; i++) {
10         if (ch[0][i] == c) {
11             return i;
12         }
13     }
14     return -1;
15 }
16 
17 int main() {
18     for (int i = 0; i < 17; i++) {    
19         if (!(i % 2)) {
20             for (int j = 0; j < 33; j++) {
21                 s[i][j] = j % 4 ? ‘-‘ : ‘+‘;
22             }
23         } else {
24             for (int j = 0; j < 33; j++) {
25                 if (j % 4) {
26                     s[i][j] = (j / 4 % 2) ^ (i / 2 % 2)? ‘:‘ : ‘.‘;
27                 } else {
28                     s[i][j] = ‘|‘;
29                 }
30             }
31         }
32         s[i][33] = ‘\0‘;
33     } 
34     for (int i = 0; i < 2; i++) { 
35         scanf("%s", r);
36         int u = r[0] == ‘B‘;
37         scanf("%s", r);
38         int n = strlen(r);
39         r[n] = ‘,‘;
40         int cun[2];
41         int now = 0, d = 5;
42         for (int j = 0; j <= n; j++) {
43             int t = type(r[j]);
44             if (r[j] == ‘,‘) {
45                 s[cun[1]][cun[0]] = ch[u][d];
46                 now = 0;
47                 d = 5; 
48             } else {
49                 if (t == -1) {
50                     cun[now] = now ? (‘8‘ - r[j]) * 2 + 1 : (r[j] - ‘a‘) * 4 + 2;
51                     now++;
52                 } else {
53                     d = t;
54                 }
55             }
56         }
57     }
58     for (int i = 0; i < 17; i++) {
59         printf("%s\n", s[i]);
60     }
61     return 0;
62 }