【HDU】3571 N-dimensional Sphere

题意：给出n维的n+1个坐标，求一个坐标，使得n+1到该坐标的距离相等。距离定义为{(x1, x2 ... xN)| ∑(xi-Xi)^2 = R^2 (i=1,2,...,N) }

首先，列出n+1个方程，显然二次项系数都为1，所以可以先把二次项消去，得到n个n元一次方程。

然后，开始高斯消元。由于java在n=50时，很久都出不了答案，c++又会爆longlong，所以我们可以选择模P。

ax=b的解等于ax=b(mod P)的解（x<P），这个很好证明。

所以在模P下，加减乘除都很容易搞定。除法用乘以逆元，乘法用二分加法，拓展欧几里德部分的乘法是不会溢出的。

为了方便处理，把输入所有的值加上一个偏移量。

  1 #include<cstdio>

  2 #include<cstring>

  3 #include<algorithm>

  4 typedef long long LL;

  5 #define MAXN 60

  6 #define P 200000000000000003LL

  7 #define S 100000000000000000LL

  8 using namespace std;

  9 LL x[MAXN], g[MAXN][MAXN], a[MAXN][MAXN], b[MAXN][MAXN];

 10 int n;

 11 inline LL Mod(LL x) {

 12     if (x >= P)

 13         return x - P;

 14     return x;

 15 }

 16 LL MulMod(LL a, LL b) {

 17     LL res;

 18     for (res = 0; b; b >>= 1) {

 19         if (b & 1)

 20             res = Mod(res + a);

 21         a = Mod(a + a);

 22     }

 23     return res;

 24 }

 25 LL ExtGcd(LL a, LL b, LL &x, LL &y) {

 26     if (b == 0) {

 27         x = 1;

 28         y = 0;

 29         return a;

 30     }

 31     LL t, d;

 32     d = ExtGcd(b, a % b, x, y);

 33     t = x;

 34     x = y;

 35     y = t - a / b * y;

 36     return d;

 37 }

 38 LL InvMod(LL a, LL n) {

 39     LL x, y;

 40     ExtGcd(a, n, x, y);

 41     return (x % n + n) % n;

 42 }

 43 void Gauss() {

 44     int i, j, k;

 45     LL inv, 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 (i != j) {

 52             for (k = i; k <= n; k++)

 53                 swap(g[i][k], g[j][k]);

 54         }

 55         inv = InvMod(g[i][i], P);

 56         for (j = i + 1; j < n; j++) {

 57             if (g[j][i]) {

 58                 tmp = MulMod(g[j][i], inv);

 59                 for (k = i; k <= n; k++) {

 60                     g[j][k] -= MulMod(tmp, g[i][k]);

 61                     g[j][k] = (g[j][k] % P + P) % P;

 62                 }

 63             }

 64         }

 65     }

 66     for (i = n - 1; i >= 0; i--) {

 67         tmp = 0;

 68         for (j = i + 1; j < n; j++) {

 69             tmp += MulMod(x[j], g[i][j]);

 70             if (tmp >= P)

 71                 tmp -= P;

 72         }

 73         tmp = g[i][n] - tmp;

 74         tmp = (tmp % P + P) % P;

 75         x[i] = MulMod(tmp, InvMod(g[i][i], P));

 76     }

 77 }

 78 int main() {

 79     int c, ca = 1;

 80     int i, j;

 81     LL tmp;

 82     scanf("%d", &c);

 83     while (c--) {

 84         scanf("%d", &n);

 85         memset(g, 0, sizeof(g));

 86         memset(b, 0, sizeof(b));

 87         for (i = 0; i <= n; i++) {

 88             for (j = 0; j < n; j++) {

 89                 scanf("%I64d", &a[i][j]);

 90                 a[i][j] += S;

 91                 b[i][n] += MulMod(a[i][j], a[i][j]);

 92                 if (b[i][n] >= P)

 93                     b[i][n] -= P;

 94             }

 95         }

 96         for (i = 0; i < n; i++) {

 97             for (j = 0; j < n; j++) {

 98                 tmp = a[i + 1][j] - a[i][j];

 99                 tmp = (tmp % P + P) % P;

100                 g[i][j] = MulMod(tmp, 2);

101             }

102             g[i][n] = b[i + 1][n] - b[i][n];

103             g[i][n] = (g[i][n] % P + P) % P;

104         }

105         Gauss();

106         printf("Case %d:\n", ca++);

107         printf("%I64d", x[0] - S);

108         for (i = 1; i < n; i++)

109             printf(" %I64d", x[i] - S);

110         putchar('\n');

111     }

112     return 0;

113 }