http://acm.hdu.edu.cn/showproblem.php?pid=3571
比赛前一天晚上刚跟haozi研究了最小包围球，其中要根据四个点求出球心坐标，然后我提出了这么拓展到N维的情况。今天比赛的时候居然就出了这么一题，虽然知道可以用高斯消元，但是这题的数据好大，有10 17那么大，而且要是整数解，用double精度根本不够。我想利用大整数，要队友写了一个java的，每次消元的时候取最小公倍数，结果TLE了。。。比赛的时候只有haozi他们队和电子科大的一个队伍过了，Orz！！！
赛后，问了一下haozi，因为最后的答案在[ -10 17 , 10 17] 的范围内，可以取一个大于2×10 17的素数p，计算的过程中都 mod p，由于坐标有正有负，可以先将球心坐标都加上10 17 这么一个偏移量，最后求出答案之后在剪掉。
PS: 高斯消元改自浙大模板
比赛前一天晚上刚跟haozi研究了最小包围球，其中要根据四个点求出球心坐标，然后我提出了这么拓展到N维的情况。今天比赛的时候居然就出了这么一题，虽然知道可以用高斯消元，但是这题的数据好大，有10 17那么大，而且要是整数解，用double精度根本不够。我想利用大整数，要队友写了一个java的，每次消元的时候取最小公倍数，结果TLE了。。。比赛的时候只有haozi他们队和电子科大的一个队伍过了，Orz！！！
赛后，问了一下haozi，因为最后的答案在[ -10 17 , 10 17] 的范围内，可以取一个大于2×10 17的素数p，计算的过程中都 mod p，由于坐标有正有负，可以先将球心坐标都加上10 17 这么一个偏移量，最后求出答案之后在剪掉。
PS: 高斯消元改自浙大模板
1
#include
<
cstdio
>
2 #include < iostream >
3 #include < map >
4 #include < queue >
5 #include < complex >
6 #include < algorithm >
7 #include < cmath >
8 #include < cstring >
9 using namespace std;
10 typedef __int64 ll;
11
12 const ll p = 1000000000000000003LL;
13 const int maxn = 60 ;
14 const ll inf = 100000000000000000LL;
15
16 ll mod( ll a, ll n ) { return ( a % n + n ) % n; }
17
18 ll mul_mod ( ll a, ll b )
19 {
20 ll ret = 0 , tmp = a % p;
21 while ( b )
22 {
23 if ( b & 1 )
24 if ( ( ret += tmp ) >= p )
25 ret -= p;
26 if ( ( tmp <<= 1 ) >= p ) tmp -= p;
27 b >>= 1 ;
28 }
29 return ret;
30 }
31
32 void gcd( ll a, ll b, ll & d, ll & x, ll & y )
33 {
34 if ( ! b ) { d = a; x = 1 ; y = 0 ; }
35 else
36 {
37 gcd( b, a % b, d, y, x );
38 y -= x * ( a / b );
39 }
40 }
41
42 ll inv( ll a, ll n )
43 {
44 ll d, x, y;
45 gcd( a, p, d, x, y );
46 return d == 1 ? mod( x, p ) : - 1 ;
47 }
48
49 ll ABS( ll a ) { return ( a < 0 ? - a : a ); }
50
51 int Gauss( int n, ll a[][maxn], ll b[] )
52 {
53 int i, j, k, row;
54 ll maxp, t;
55 for ( k = 0 ; k < n; k ++ )
56 {
57 for ( maxp = 0 , i = k; i < n; i ++ )
58 if ( ABS( a[i][k] ) > ABS( maxp ) )
59 maxp = a[row = i][k];
60 if ( maxp == 0 ) return 0 ;
61 if ( row != k )
62 {
63 for ( j = k; j < n; j ++ )
64 t = a[k][j], a[k][j] = a[row][j], a[row][j] = t;
65 t = b[k], b[k] = b[row], b[row] = t;
66 }
67 ll INV = inv( maxp, p );
68 for ( j = k + 1 ; j < n; j ++ )
69 {
70 a[k][j] = mul_mod( a[k][j], INV );
71 for ( i = k + 1 ; i < n; i ++ )
72 a[i][j] = mod( a[i][j] - mul_mod( a[i][k], a[k][j] ), p );
73 }
74 b[k] = mul_mod( b[k], INV );
75 for ( i = k + 1 ; i < n; i ++ )
76 b[i] = mod( b[i] - mul_mod( b[k], a[i][k] ), p );
77 }
78 for ( i = n - 1 ; i >= 0 ; i -- )
79 for ( j = i + 1 ; j < n; j ++ )
80 b[i] = mod( b[i] - mul_mod( a[i][j], b[j] ), p );
81 return 1 ;
82 }
83
84 int main( int argc, char * argv[])
85 {
86 int cas, n;
87 ll a[maxn][maxn], b[maxn], c[maxn][maxn], d[maxn];
88 scanf( " %d " , & cas);
89 for ( int t = 1 ; t <= cas; t ++ )
90 {
91 scanf( " %d " , & n);
92 for ( int i = 0 ; i <= n; i ++ )
93 {
94 b[i] = 0 ;
95 for ( int j = 0 ; j < n ; j ++ )
96 {
97 scanf( " %I64d " , & a[i][j]);
98 a[i][j] += inf;
99 b[i] = ( b[i] + mul_mod( a[i][j], a[i][j] ) ) % p;
100 a[i][j] = ( a[i][j] + a[i][j] ) % p;
101 }
102 }
103 for ( int i = 0 ; i < n; i ++ )
104 {
105 for ( int j = 0 ; j < n; j ++ )
106 c[i][j] = mod( a[i][j] - a[n][j], p );
107 d[i] = mod( b[i] - b[n], p );
108 }
109 Gauss( n, c, d );
110 // gauss_cpivot( n, c, d );
111 printf( " Case %d:\n " ,t);
112 printf( " %I64d " ,d[ 0 ] - inf);
113 for ( int i = 1 ; i < n; i ++ )
114 printf( " %I64d " ,d[i] - inf);
115 printf( " \n " );
116 }
117 return 0 ;
118 }
2 #include < iostream >
3 #include < map >
4 #include < queue >
5 #include < complex >
6 #include < algorithm >
7 #include < cmath >
8 #include < cstring >
9 using namespace std;
10 typedef __int64 ll;
11
12 const ll p = 1000000000000000003LL;
13 const int maxn = 60 ;
14 const ll inf = 100000000000000000LL;
15
16 ll mod( ll a, ll n ) { return ( a % n + n ) % n; }
17
18 ll mul_mod ( ll a, ll b )
19 {
20 ll ret = 0 , tmp = a % p;
21 while ( b )
22 {
23 if ( b & 1 )
24 if ( ( ret += tmp ) >= p )
25 ret -= p;
26 if ( ( tmp <<= 1 ) >= p ) tmp -= p;
27 b >>= 1 ;
28 }
29 return ret;
30 }
31
32 void gcd( ll a, ll b, ll & d, ll & x, ll & y )
33 {
34 if ( ! b ) { d = a; x = 1 ; y = 0 ; }
35 else
36 {
37 gcd( b, a % b, d, y, x );
38 y -= x * ( a / b );
39 }
40 }
41
42 ll inv( ll a, ll n )
43 {
44 ll d, x, y;
45 gcd( a, p, d, x, y );
46 return d == 1 ? mod( x, p ) : - 1 ;
47 }
48
49 ll ABS( ll a ) { return ( a < 0 ? - a : a ); }
50
51 int Gauss( int n, ll a[][maxn], ll b[] )
52 {
53 int i, j, k, row;
54 ll maxp, t;
55 for ( k = 0 ; k < n; k ++ )
56 {
57 for ( maxp = 0 , i = k; i < n; i ++ )
58 if ( ABS( a[i][k] ) > ABS( maxp ) )
59 maxp = a[row = i][k];
60 if ( maxp == 0 ) return 0 ;
61 if ( row != k )
62 {
63 for ( j = k; j < n; j ++ )
64 t = a[k][j], a[k][j] = a[row][j], a[row][j] = t;
65 t = b[k], b[k] = b[row], b[row] = t;
66 }
67 ll INV = inv( maxp, p );
68 for ( j = k + 1 ; j < n; j ++ )
69 {
70 a[k][j] = mul_mod( a[k][j], INV );
71 for ( i = k + 1 ; i < n; i ++ )
72 a[i][j] = mod( a[i][j] - mul_mod( a[i][k], a[k][j] ), p );
73 }
74 b[k] = mul_mod( b[k], INV );
75 for ( i = k + 1 ; i < n; i ++ )
76 b[i] = mod( b[i] - mul_mod( b[k], a[i][k] ), p );
77 }
78 for ( i = n - 1 ; i >= 0 ; i -- )
79 for ( j = i + 1 ; j < n; j ++ )
80 b[i] = mod( b[i] - mul_mod( a[i][j], b[j] ), p );
81 return 1 ;
82 }
83
84 int main( int argc, char * argv[])
85 {
86 int cas, n;
87 ll a[maxn][maxn], b[maxn], c[maxn][maxn], d[maxn];
88 scanf( " %d " , & cas);
89 for ( int t = 1 ; t <= cas; t ++ )
90 {
91 scanf( " %d " , & n);
92 for ( int i = 0 ; i <= n; i ++ )
93 {
94 b[i] = 0 ;
95 for ( int j = 0 ; j < n ; j ++ )
96 {
97 scanf( " %I64d " , & a[i][j]);
98 a[i][j] += inf;
99 b[i] = ( b[i] + mul_mod( a[i][j], a[i][j] ) ) % p;
100 a[i][j] = ( a[i][j] + a[i][j] ) % p;
101 }
102 }
103 for ( int i = 0 ; i < n; i ++ )
104 {
105 for ( int j = 0 ; j < n; j ++ )
106 c[i][j] = mod( a[i][j] - a[n][j], p );
107 d[i] = mod( b[i] - b[n], p );
108 }
109 Gauss( n, c, d );
110 // gauss_cpivot( n, c, d );
111 printf( " Case %d:\n " ,t);
112 printf( " %I64d " ,d[ 0 ] - inf);
113 for ( int i = 1 ; i < n; i ++ )
114 printf( " %I64d " ,d[i] - inf);
115 printf( " \n " );
116 }
117 return 0 ;
118 }