【POJ】2065 SETI

题意：直接拿样例，37 abc。

a～z表示1～26，*表示0。

x0*1^0+x1*1^1+x2*1^2=1(mod 37)

x0*2^0+x1*2^1+x2*2^2=2(mod 37)

x0*3^0+x1*3^1+x2*3^2=3(mod 37)

高斯消元，除法x等于乘以x对p的逆元。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define MAXN 110
 5 using namespace std;
 6 char str[MAXN];
 7 int n, p;
 8 int g[MAXN][MAXN], x[MAXN];
 9 int GET(char ch) {
10     if (ch == '*')
11         return 0;
12     return ch - 'a' + 1;
13 }
14 int PowMod(int a, int b, int c) {
15     int res;
16     for (res = 1; b; b >>= 1) {
17         if (b & 1) {
18             res *= a;
19             res %= c;
20         }
21         a *= a;
22         a %= c;
23     }
24     return res;
25 }
26 int EXTGCD(int a, int b, int &x, int &y) {
27     int t, d;
28     if (b == 0) {
29         x = 1;
30         y = 0;
31         return a;
32     }
33     d = EXTGCD(b, a % b, x, y);
34     t = x;
35     x = y;
36     y = t - a / b * y;
37     return d;
38 }
39 int Invmod(int a, int n) {
40     int x, y;
41     EXTGCD(a, n, x, y);
42     return (x % n + n) % n;
43 }
44 void Gauss() {
45     int i, j, k, tmp;
46     for (i = 0; i < n; i++) {
47         for (j = i; j < n; j++) {
48             if (g[j][i])
49                 break;
50         }
51         if (j != i) {
52             for (k = i; k <= n; k++)
53                 swap(g[j][k], g[i][k]);
54         }
55         for (j = i + 1; j < n; j++) {
56             if (g[j][i]) {
57                 tmp = (g[j][i] * Invmod(g[i][i], p)) % p;
58                 for (k = i; k <= n; k++) {
59                     g[j][k] -= tmp * g[i][k];
60                     g[j][k] = (g[j][k] % p + p) % p;
61                 }
62             }
63         }
64     }
65     for (i = n - 1; i >= 0; i--) {
66         tmp = 0;
67         for (j = i + 1; j < n; j++) {
68             tmp += x[j] * g[i][j];
69             tmp = (tmp % p + p) % p;
70         }
71         tmp = g[i][n] - tmp;
72         tmp = (tmp % p + p) % p;
73         x[i] = tmp * Invmod(g[i][i], p) % p;
74     }
75 }
76 int main() {
77     int c;
78     int i, j;
79     scanf("%d", &c);
80     while (c--) {
81         scanf("%d %s", &p, str);
82         n = strlen(str);
83         for (i = 0; i < n; i++) {
84             g[i][n] = GET(str[i]);
85             for (j = 0; j < n; j++)
86                 g[i][j] = PowMod(i + 1, j, p);
87         }
88         Gauss();
89         printf("%d", x[0]);
90         for (i = 1; i < n; i++)
91             printf(" %d", x[i]);
92         putchar('\n');
93     }
94     return 0;
95 }