UVA 1563，uva1563

UVA 1563，uva1563
UVA 1563 - SETI

題目連結

題意：根據題目那個式子，構造一個序列，能產生相應字串

思路：根據式子能構造出n個方程，一共解n個未知量，利用高斯消元去解，中間過程有取摸過程，所以遇到除法的時候要使用逆元去搞

代碼：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 105;int pow_mod(int x, int k, int mod) {    int ans = 1;    while (k) {if (k&1) ans = ans * x % mod;x = x * x % mod;k >>= 1;    }    return ans;}int inv(int a, int n) {    return pow_mod(a, n - 2, n);}int t, p, n, A[N][N];char str[N];int hash(int c) {    if (c == '*') return 0;    return c - 'a' + 1;}void build() {    for (int i = 0; i < n; i++) {A[i][n] = hash(str[i]);int tmp = 1;for (int j = 0; j < n; j++) {    A[i][j] = tmp;    tmp = tmp * (i + 1) % p;}    }}void gauss() {    for (int i = 0; i < n; i++) {int r;for (r = i; r < n; i++)    if (A[r][i]) break;if (r == n) continue;for (int j = i; j <= n; j++) swap(A[r][j], A[i][j]);for (int j = 0; j < n; j++) {    if (i == j) continue;    if (A[j][i]) {int tmp = A[j][i] * inv(A[i][i], p) % p;for (int k = i; k <= n; k++) {    A[j][k] = (((A[j][k] - tmp * A[i][k]) % p) + p) % p;}    }}    }    for (int i = 0; i < n; i++)printf("%d%c", A[i][n] * inv(A[i][i], p) % p, i == n - 1 ? '\n' : ' ');}int main() {    scanf("%d", &t);    while (t--) {scanf("%d%s", &p, str);n = strlen(str);build();gauss();    }    return 0;}