同余方程组、拓展中国剩余定理 excrt
公式：\(x \equiv b_i(\bmod\ a_i)\) ，即 \((x - b_i) \mid a_i\) 。
int n; LL ai[maxn], bi[maxn];
inline int mypow(int n, int k, int p) {int r = 1;for (; k; k >>= 1, n = n * n % p)if (k & 1) r = r * n % p;return r;
}
LL exgcd(LL a, LL b, LL &x, LL &y) {if (b == 0) { x = 1, y = 0; return a; }LL gcd = exgcd(b, a % b, x, y), tp = x;x = y, y = tp - a / b * y;return gcd;
}
LL excrt() {LL x, y, k;LL M = bi[1], ans = ai[1];for (int i = 2; i <= n; ++ i) {LL a = M, b = bi[i], c = (ai[i] - ans % b + b) % b;LL gcd = exgcd(a, b, x, y), bg = b / gcd;if (c % gcd != 0) return -1;x = mul(x, c / gcd, bg);ans += x * M;M *= bg;ans = (ans % M + M) % M;}return (ans % M + M) % M;
}
int main() {cin >> n;for (int i = 1; i <= n; ++ i) cin >> bi[i] >> ai[i];cout << excrt() << endl;return 0;
}
求解连续按位异或
以 \(\mathcal O(1)\) 复杂度计算 \(0\oplus1\oplus\dots\oplus n\) 。
unsigned xor_n(unsigned n) {unsigned t = n & 3;if (t & 1) return t / 2u ^ 1;return t / 2u ^ n;
}
i64 xor_n(i64 n) {if (n % 4 == 1) return 1;else if (n % 4 == 2) return n + 1;else if (n % 4 == 3) return 0;else return n;
}
高斯消元求解线性方程组
题目大意：输入一个包含 \(N\) 个方程 \(N\) 个未知数的线性方程组，系数与常数均为实数（两位小数）。求解这个方程组。如果存在唯一解，则输出所有 \(N\) 个未知数的解，结果保留两位小数。如果无数解，则输出 \(\tt{}X\) ，如果无解，则输出 \(\tt{}N\) 。
const int N = 110;
const double eps = 1e-8;
LL n;
double a[N][N];
LL gauss(){LL c, r;for (c = 0, r = 0; c < n; c ++ ){LL t = r;for (int i = r; i < n; i ++ ) //找到绝对值最大的行 if (fabs(a[i][c]) > fabs(a[t][c]))t = i;if (fabs(a[t][c]) < eps) continue;for (int j = c; j < n + 1; j ++ ) swap(a[t][j], a[r][j]); //将绝对值最大的一行换到最顶端for (int j = n; j >= c; j -- ) a[r][j] /= a[r][c]; //将当前行首位变成 1for (int i = r + 1; i < n; i ++ ) //将下面列消成 0 if (fabs(a[i][c]) > eps)for (int j = n; j >= c; j -- )a[i][j] -= a[r][j] * a[i][c];r ++ ;}if (r < n){for (int i = r; i < n; i ++ )if (fabs(a[i][n]) > eps)return 2;return 1;}for (int i = n - 1; i >= 0; i -- )for (int j = i + 1; j < n; j ++ )a[i][n] -= a[i][j] * a[j][n];return 0;
}
int main(){cin >> n;for (int i = 0; i < n; i ++ )for (int j = 0; j < n + 1; j ++ )cin >> a[i][j];LL t = gauss();if (t == 0){for (int i = 0; i < n; i ++ ){if (fabs(a[i][n]) < eps) a[i][n] = abs(a[i][n]);printf("%.2lf\n", a[i][n]);}}else if (t == 1) cout << "Infinite group solutions\n";else cout << "No solution\n";return 0;
}