学步园

题目链接：uva 1563 - SETI

题目大意：给定p和一个字符串，求一个序列，对应的f(k)根据公式求出的值为字符串上第k个字符（‘*’对应值为0，从a到z对应1到26）

解题思路：根据公式列出模方程组，然后高斯消元求解。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 80;
typedef int Mat[maxn][maxn];

int p, n;
char s[maxn];
Mat A;

inline int cat (char ch) {
    if (ch == '*')
        return 0;
    return ch - 'a' + 1;
}

void solve () {
    n = strlen(s);
    memset(A, 0, sizeof(A));
    for (int i = 0; i < n; i++) {
        int k = 1;
        for (int j = 0; j < n; j++) {
            A[i][j] = k;
            k = k * (i+1) % p;
        }
        A[i][n] = cat(s[i]) % p;
    }
}

int pow_mod(int x, int k, int mod) {
    int ret = 1;
    while (k) {
        if (k&1)
            ret = ret * x % mod;
        x = x * x % mod;
        k >>= 1;
    }
    return ret;
}

int inv (int v, int mod) {
    return pow_mod(v % mod, mod-2, mod);
}


void gauss_elimin () {
    for (int i = 0; i < n; i++) {
        int r = i;

        for (int j = i+1; j < n; j++)
            if (A[r][i] < A[j][i])
                r = j;

        if (r != i) {
            for (int j = 0; j <= n; j++)
                swap(A[i][j], A[r][j]);
        }

        if (A[i][i] == 0)
            continue;

        for (int k = i + 1; k < n; k++) {
            int f = A[k][i] * inv(A[i][i], p) % p;
            for (int j = i; j <= n; j++)
                A[k][j] = ((A[k][j] - A[i][j] * f) % p + p) % p;
        }
    }

    for (int i = n-1; i >= 0; i--) {
        for (int j = i+1; j < n; j++)
            A[i][n] = ((A[i][n] - A[i][j] * A[j][n]) % p + p) % p;
        A[i][n] = A[i][n] * inv(A[i][i], p) % p;
    }

    printf("%d", A[0][n]);
    for (int i = 1; i < n; i++)
        printf(" %d", A[i][n]);
    printf("\n");
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d%s", &p, s);
        solve();
        gauss_elimin();
    }
    return 0;
}