题目原意很简单，就是解一个三元一次方程组
直接高斯消元解方程组，枚举最后一列的倍数(k)
注意double的精度，有很多细节需要处理
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 else
103 {
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; //line1
128 cin>>A2>>B2>>C2; //line2
129 cin>>A3>>B3>>C3; //line3
130
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 else
158 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 */
时间： 2024-10-10 09:58:58