bzoj1013 [JSOI2008]球形空间产生器sphere

题目链接

第一次写高斯消元

注意考虑分母、分子为零的情况

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<string>
 7 #include<cmath>
 8 #include<ctime>
 9 #include<queue>
10 #include<stack>
11 #include<map>
12 #include<set>
13 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
14 #define re(i,l,r) for(int i=(l);i<=(r);i++)
15 #define Clear(a,b) memset(a,b,sizeof(a))
16 #define inout(x) printf("%d",(x))
17 #define douin(x) scanf("%lf",&x)
18 #define strin(x) scanf("%s",(x))
19 #define LLin(x) scanf("%lld",&x)
20 #define op operator
21 #define CSC main
22 typedef unsigned long long ULL;
23 typedef const int cint;
24 typedef long long LL;
25 using namespace std;
26 double f(const double &a){return a*a;}
27 const double eps=1e-8;
28 void inin(int &ret)
29 {
30     ret=0;int f=0;char ch=getchar();
31     while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();}
32     while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar();
33     ret=f?-ret:ret;
34 }
35 int n;
36 double xl[11],temp;
37 double a[11][11],b[11],x[11];
38 int main()
39 {
40     freopen("in.in","r",stdin);
41     freopen("out.out","w",stdout);
42     inin(n);
43     re(i,1,n)douin(xl[i]);
44     re(i,1,n)re(j,1,n)
45     {
46         douin(temp);
47         a[i][j]=2*(temp-xl[j]);
48         a[i][n+1]+=f(temp)-f(xl[j]);
49     }
50     re(i,1,n-1)
51     {
52         if(abs(a[i][i]<eps))
53         re(j,i+1,n)if(abs(a[j][i])>eps)
54         {
55             re(k,1,n+1)a[i][k]+=a[j][k];
56             break;
57         }
58         re(j,i+1,n)
59         {
60             if(abs(a[j][i])<eps)continue;
61             double bi=a[i][i]/a[j][i];
62             re(k,i,n+1)a[j][k]*=bi;
63             re(k,i,n+1)a[j][k]-=a[i][k];
64         }
65     }
66     rre(i,n,1)
67     {
68         x[i]=a[i][n+1]/a[i][i];
69         rre(j,i-1,1)a[j][n+1]-=a[j][i]*x[i],a[j][i]=0;
70     }
71     re(i,1,n)printf("%.3f%c",x[i],i==n?'\n':' ');
72      return 0;
73 }

 

posted @ 2016-03-08 08:52  HugeGun  阅读(191)  评论(0编辑  收藏  举报