1 /*  2 PROB:ratios  3 LANG:C++  4 */  5   6 #include <stdio.h>  7 #include <math.h>  8 #include <algorithm>  9 #include <iostream> 10 using namespace std; 11 #define maxn 5 12 double A[5][5],a[5][5]; 13  14 typedef double Matrix[maxn][maxn]; 15  16 void solve(Matrix A,int n) 17 { 18     int i,j,k,r; 19  20     for (i=0;i<n;i++) 21     { 22         r=i; 23         for (j=i+1;j<n;j++) 24             if (fabs(A[j][i])>fabs(A[r][i])) 25                 r=j; 26         if (r!=i) 27             for (j=0;j<=n;j++) 28                 swap(A[r][j],A[i][j]); 29  30         for (k=i+1;k<n;k++) 31         { 32             double f=A[k][i]/A[i][i]; 33             for (j=i;j<=n;j++) 34                 A[k][j]-=f*A[i][j]; 35         } 36     } 37     for (i=n-1;i>=0;i--) 38     { 39         for (j=i+1;j<n;j++) 40             A[i][n]-=A[j][n]*A[i][j]; 41         A[i][n]/=A[i][i]; 42     } 43 } 44  45 bool satisify(double x,double y,double z) 46 { 47     int t1=x+0.5,t2=y+0.5,t3=z+0.5;     //+0.5四舍五入，处理精度问题 48                                         //比如原来是4.9999999，直接转成int就成了4，完蛋啦T^T 49     double xx=x-t1,yy=y-t2,zz=z-t3; 50     double comp=pow(10,-5);             //double的比较法= =，差值小于10^-10就算相等了 51     if ((fabs(xx)<=comp)&&(fabs(yy)<=comp)&&(fabs(zz)<=comp)) 52         return true; 53     else 54         return false; 55 } 56  57 int main() 58 { 59     freopen("ratios.in","r",stdin); 60     freopen("ratios.out","w",stdout); 61  62     int x,y,z,t1,t2,t3; 63     double dx,dy,dz; 64     bool sol=false; 65  66     scanf("%d %d %d",&x,&y,&z); 67     //A[0][3]=x;  A[1][3]=y;  A[2][3]=z; 68     t1=x;   t2=y;   t3=z; 69     scanf("%d %d %d",&x,&y,&z); 70     A[0][0]=x;  A[1][0]=y;  A[2][0]=z; 71     scanf("%d %d %d",&x,&y,&z); 72     A[0][1]=x;  A[1][1]=y;  A[2][1]=z; 73     scanf("%d %d %d",&x,&y,&z); 74     A[0][2]=x;  A[1][2]=y;  A[2][2]=z; 75 /* 76     for (int i=0;i<=2;i++) 77     { 78         for (int j=0;j<=3;j++) 79             printf("%.8f ",A[i][j]); 80         printf("\n"); 81     } 82 */ 83     for (int i=1;i<=100;i++) 84     { 85         for (int i=0;i<=2;i++) 86             for (int j=0;j<=3;j++) 87                 a[i][j]=A[i][j]; 88         a[0][3]=t1*i; 89         a[1][3]=t2*i; 90         a[2][3]=t3*i; 91         //printf("%.8f %.8f %.8f\n",a[0][3],a[1][3],a[2][3]); 92         solve(a,3); 93         dx=a[0][3];     dy=a[1][3];     dz=a[2][3]; 94         //printf("%.8f %.8f %.8f\n",dx,dy,dz); 95         if (satisify(dx,dy,dz)) 96         { 97             sol=true; 98             int xx=dx,yy=dy,zz=dz; 99             //printf("%d %d %d\n",xx,yy,zz);100             if (xx<0||yy<0||zz<0)101                 printf("NONE\n");102             else103             {104                 xx=dx+0.5;  yy=dy+0.5;  zz=dz+0.5;105                 printf("%d %d %d %d\n",xx,yy,zz,i);106             }107             break;108         }109         //printf("%.8f %.8f %.8f\n",a[0][3],a[1][3],a[2][3]);110     }111     if (!sol) printf("NONE\n");112 113 }114 115 /*116 117 118 int main()              //垃圾代码，一开始YY的错了119 {120     freopen("ratios.in","r",stdin);121     freopen("ratios.out","w",stdout);122 123     double a1,a2,a3,a0,b1,b2,b3,b0,c1,c2,c3,c0,A0,A1,A2,A3,B0,B1,B2,B3,C0,C1,C2,C3;124     int xx,yy,zz;125     bool sol=false;126     cin>>A0>>B0>>C0;127     cin>>A1>>B1>>C1;    //line1128     cin>>A2>>B2>>C2;    //line2129     cin>>A3>>B3>>C3;    //line3130 131 for (int i=1;i<=100;i++)132 {133     a0=A0*i;  a1=A1;  a2=A2;  a3=A3;134     b0=B0*i;  b1=B1;  b2=B2;  b3=B3;135     c0=C0*i;  c1=C1;  c2=C2;  c3=C3;136 137     double t0=b1*c1,t1=c1*a1,t2=a1*b1;138     a1=a1*t0;   a2=a2*t0;   a3=a3*t0;   a0=a0*t0;139     b1=b1*t1;   b2=b2*t1;   b3=b3*t1;   b0=b0*t1;140     c1=c1*t2;   c2=c2*t2;   c3=c3*t2;   c0=c0*t2;141     c1=c1-a1;   c2=c2-a2;   c3=c3-a3;   c0=c0-a0;142     b1=b1-a1;   b2=b2-a2;   b3=b3-a3;   b0=b0-a0;143     t0=c2;      t1=b2;144     b1=b1*t0;   b2=b2*t0;   b3=b3*t0;   b0=b0*t0;145     c1=c1*t1;   c2=c2*t1;   c3=c3*t1;   c0=c0*t1;146     c1=c1-b1;   c2=c2-b2;   c3=c3-b3;   c0=c0-b0;147     double z=c0/c3;148     double y=(b0-b3*z)/b2;149     double x=(a0-a3*z-a2*y)/a1;150     printf("%%f %f %f\n",x,y,z);151     if (satisify(x,y,z))152     {153         sol=true;154         xx=x,yy=y,zz=z;155         if (xx<0||yy<0||zz<0)156             printf("NONE\n");157         else158             printf("%d %d %d %d\n",xx,yy,zz,i);159         break;160     }161 }162     if (!sol) printf("NONE\n");163     return 0;164 }165 */