洛谷P3389 【模板】高斯消元法

传送门

 

 1 //minamoto
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cmath>
 5 using namespace std;
 6 #define getc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
 7 char buf[1<<21],*p1=buf,*p2=buf;
 8 inline int read(){
 9     #define num ch-'0'
10     char ch;bool flag=0;int res;
11     while((ch=getc())>'9'||ch<'0')
12     (ch=='-')&&(flag=true);
13     for(res=num;(ch=getc())<='9'&&ch>='0';res=res*10+num);
14     (flag)&&(res=-res);
15     #undef num
16     return res;
17 }
18 const int N=105;
19 double mp[N][N],ans[N],eps=1e-7;
20 int main(){
21 //    freopen("testdata.in","r",stdin);
22     int n=read();
23     for(int i=1;i<=n;++i)
24     for(int j=1;j<=n+1;++j)
25     mp[i][j]=read();
26     for(int i=1;i<=n;++i){
27         int r=i;
28         for(int j=i+1;j<=n;++j)
29         if(fabs(mp[r][i])<fabs(mp[j][i])) r=j;
30         if(fabs(mp[r][i])<eps) return puts("No Solution"),0;
31         if(i!=r) swap(mp[i],mp[r]);
32         double div=mp[i][i];
33         for(int j=i;j<=n+1;++j) mp[i][j]/=div;
34         for(int j=i+1;j<=n;++j){
35             div=mp[j][i];
36             for(int k=i;k<=n+1;++k)
37             mp[j][k]-=mp[i][k]*div;
38         }
39     }
40     ans[n]=mp[n][n+1];
41     for(int i=n-1;i;--i){
42         ans[i]=mp[i][n+1];
43         for(int j=i+1;j<=n;++j)
44         ans[i]-=mp[i][j]*ans[j];
45     }
46     for(int i=1;i<=n;++i) printf("%.2lf\n",ans[i]);
47     return 0;
48 }

 

posted @ 2018-09-30 13:33  bztMinamoto  阅读(162)  评论(0)  编辑  收藏  举报
Live2D