链接：
http://poj.org/problem?id=2420
题意：
给出n个点，找到一个点，使得它到所有的点的距离最小。
题解：
最近要做一个排课系统，需要用到模拟退火算法，之前虽然了解过这个算法，但是没有写过题。就先在POJ上找了一道学习一下。
代码：
1 #include <iomanip> 2 struct Point { double x, y; }; 3 4 const double eps = 1e-8; //搜索条件阀值 5 const double T = 100; //初始温度 6 const double delta = 0.98; //温度下降速度 7 int dx[4] = { 1,-1,0,0 }; 8 int dy[4] = { 0,0,1,-1 }; 9 int n; 10 Point p[MAXN]; 11 12 double sqr(double x) { 13 return x * x; 14 } 15 16 double dist(Point a, Point b) { 17 return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y)); 18 } 19 20 double sum(Point point) { 21 double res = 0; 22 rep(i, 0, n) res += dist(point, p[i]); 23 return res; 24 } 25 26 int main() { 27 ios::sync_with_stdio(false), cin.tie(0); 28 cin >> n; 29 rep(i, 0, n) cin >> p[i].x >> p[i].y; 30 Point s = p[0]; //随机初始化一个点开始搜索 31 double t = T; //初始化温度 32 double ans = INF; 33 while (t > eps) { 34 int fg = 1; 35 while (fg) { 36 fg = 0; 37 rep(i, 0, 4) { 38 Point point = Point{ s.x + dx[i],s.y + dy[i] }; 39 double t = sum(point); 40 if (ans > t) { 41 ans = t; 42 s = point; 43 fg = 1; 44 } 45 } 46 } 47 t *= delta; 48 } 49 cout << std::fixed << setprecision(0) << ans << endl; 50 return 0; 51 }
链接：
http://poj.org/problem?id=1808
题意：
判断x^2同余a(modn)是否存在
题解：
平方剩余
代码：
31 ll mod_pow(ll x, ll n, ll mod) { 32 int res = 1; 33 while (n) { 34 if (n & 1) res = res * x % mod; 35 x = x * x % mod; 36 n >>= 1; 37 } 38 return res; 39 } 40 41 ll mod_sqr(ll a, ll n) { 42 ll b, k, i, x; 43 //if (n == 2) return a%n; 44 if(mod_pow(a, (n - 1) / 2, n) == 1) { 45 return 1; 46 if (n % 4 == 3) x = mod_pow(a, (n + 1) / 4, n); 47 else { 48 for (b = 1; mod_pow(b, (n - 1) / 2, n) == 1; b++); 49 i = (n - 1) / 2; 50 k = 0; 51 do { 52 i /= 2; 53 k /= 2; 54 if ((mod_pow(a, i, n)*mod_pow(b, k, n) + 1) % n == 0) 55 k += (n - 1) / 2; 56 } while (i % 2 == 0); 57 x = mod_pow(a, (i + 1) / 2, n)*mod_pow(b, k / 2, n) % n; 58 } 59 if (x * 2 > n) x = n - x; 60 } 61 return -1; 62 } 63 64 int main() { 65 ios::sync_with_stdio(false), cin.tie(0); 66 int T; 67 cin >> T; 68 rep(cas, 1, T + 1) { 69 ll a, p; 70 cin >> a >> p; 71 a = (a%p + p) % p; 72 cout << "Scenario #" << cas << ":" << endl; 73 cout << mod_sqr(a, p) << endl << endl; 74 } 75 return 0; 76 }
链接：
http://poj.org/problem?id=1808
题意：
判断x^2同余a(modn)是否存在
题解：
平方剩余
代码：
31 ll mod_pow(ll x, ll n, ll mod) { 32 int res = 1; 33 while (n) { 34 if (n & 1) res = res * x % mod; 35 x = x * x % mod; 36 n >>= 1; 37 } 38 return res; 39 } 40 41 ll mod_sqr(ll a, ll n) { 42 ll b, k, i, x; 43 //if (n == 2) return a%n; 44 if(mod_pow(a, (n - 1) / 2, n) == 1) { 45 return 1; 46 if (n % 4 == 3) x = mod_pow(a, (n + 1) / 4, n); 47 else { 48 for (b = 1; mod_pow(b, (n - 1) / 2, n) == 1; b++); 49 i = (n - 1) / 2; 50 k = 0; 51 do { 52 i /= 2; 53 k /= 2; 54 if ((mod_pow(a, i, n)*mod_pow(b, k, n) + 1) % n == 0) 55 k += (n - 1) / 2; 56 } while (i % 2 == 0); 57 x = mod_pow(a, (i + 1) / 2, n)*mod_pow(b, k / 2, n) % n; 58 } 59 if (x * 2 > n) x = n - x; 60 } 61 return -1; 62 } 63 64 int main() { 65 ios::sync_with_stdio(false), cin.tie(0); 66 int T; 67 cin >> T; 68 rep(cas, 1, T + 1) { 69 ll a, p; 70 cin >> a >> p; 71 a = (a%p + p) % p; 72 cout << "Scenario #" << cas << ":" << endl; 73 cout << mod_sqr(a, p) << endl << endl; 74 } 75 return 0; 76 }