自编双色球缩水程序（C语言）
以下是本人自编自用的双色球缩水程序，四年多来本人最多一次中过3000（离500W就一个数，28与29的差别），中过十几次200。儿子杨朗死后就再也没有用过该程序。现将其公布在天涯，以感谢天涯对本人的支持。
编程的思路是以前面的30期为数据统计依据，计算下一期要出的号码。懂C的朋友如有不解之处，可给本人发短信，共同探讨。
main()
{
int i, j, k, l, m, n, o, p,s,t,X;
int w,R, d[6], f[6],c[33], e[33], N1,N2,N3,N4,N5,N6;
int S=0, Z;
int CS[13];
int LH[13];
int a[30][6]={{5,7,10,14,17,25},
{5,8,10,15,23,26},{3,7,13,23,27,30},{7,13,17,26,32,33},
{10,11,13,16,19,30},{10,19,20,21,23,32},{2,5,11,26,30,32},
{1,2,14,23,28,29},{8,12,20,22,30,33},{2,15,19,24,31,32},
{4,10,16,23,28,30},{6,11,18,20,25,30},{3,5,12,18,21,23},
{1,2,9,10,21,31},{4,5,23,26,31,32},{1,3,12,20,21,29},
{9,16,17,18,22,27},{5,10,16,19,23,28},{1,13,15,17,20,30},
{9,18,19,25,28,31},{1,9,14,16,28,32},{5,7,12,14,15,20},
{2,9,16,21,30,31},{1,11,13,25,32,33},{4,5,6,25,29,30},
{11,15,18,21,27,29},{2,8,12,18,24,28},{4,9,11,20,32,33},
{4,8,12,17,20,30},{11,12,13,18,23,32}};
int b[5][6]={{11,15,18,21,27,29},{2,8,12,18,24,28},
{4,9,11,20,32,33},{4,8,12,17,20,30},{11,12,13,18,23,32}};
for(k=0;k<=32;k++)
for(i=0;i<=29;i++)
for(j=0;j<=5;j++)
if(a[i][j]==k+1) c[k]=c[k]+1;
for(k=0;k<=32;k++)
for(i=0;i<=4;i++)
for(j=0;j<=5;j++)
if(b[i][j]==k+1) e[k]=e[k]+1;
for(i=0;i<=16;i++)
{
d[0]=c[i];f[0]=e[i];N1=i+1;
for(j=1;j<=19;j++)
{ d[1]=c[j];f[1]=e[j];N2=j+1;
for(k=3;k<=24;k++)
{d[2]=c[k];f[2]=e[k];N3=k+1;
for(l=4;l<=29;l++)
{d[3]=c[l];f[3]=e[l];N4=l+1;
for(m=10;m<=31;m++)
{d[4]=c[m];f[4]=e[m];N5=m+1;
for(n=15;n<=32;n++)
{d[5]=c[n];f[5]=e[n];N6=n+1;
R=0;Z=0;X=0;
for(o=0;o<=12;o++) {CS[o]=0;LH[o]=0;}
Z=N1+N2+N3+N4+N5+N6;
R=d[0]+d[1]+d[2]+d[3]+d[4]+d[5];
X=f[0]+f[1]+f[2]+f[3]+f[4]+f[5];
for(o=0;o<=12;o++)
for(p=0;p<=5;p++)
{if(d[p]==o) CS[o]=CS[o]+1;
if(f[p]==o) LH[o]=LH[o]+1;}
if((R==25||R==27||R==33)&&N1<N2&&N2<N3&&N3<N4&&N4<N5&&N5<N6&&
(Z==88||Z==94||Z==100||Z==66)&&(X==5||X==7)&&
(N1==1||N1==3||N1==5||N1==6)&&
(N6==22||N6==24||N6==26||N6==27||N6==28||Z==31)&&
(N2<11))
{if((CS[2]+CS[4])>=1&&CS[7]==0&&CS[8]>=1&&CS[9]<=1&&CS[2]<=2&&CS[6]>=1&&CS[5]<=2&&CS[6]<=3&&CS[4]<=1&&(CS[2]>=2||CS[5]==2||CS[6]>=2)&&
LH[0]>=2&&LH[0]<=3&&LH[1]>=1&&LH[1]<=2&&(LH[2]+CS[3])>=1)
{
printf("%d %d %d %d %d %d ", N1, N2, N3, N4, N5, N6);
printf(" ");
printf("%d ",Z);
printf("%d ",R);
printf("%d%d%d%d%d%d ",d[0],d[1],d[2],d[3],d[4],d[5]);
printf("%d%d%d%d%d%d ",f[0],f[1],f[2],f[3],f[4],f[5]);
printf(" ");S++; }
}
}
}
}
}
}
}
printf("S=%d ",S);
}
