组合程序

在选取红球时，另红球一编号比红球二编号小，即可得到所有的组合，具体语句为j=i+1，代码及截图如下，因为结果数据比较庞大，所以只截取开头和结尾

#include<iostream>

#include<iomanip>

using namespace std

void main()

{

int count=0

cout<<"组合种类  红球一  红球二  蓝球"<<endl

for(int i=1 i<=8 i++)

for(int j=i+1 j<=8 j++)

for(int k=1 k<=10 k++)

{

cout<<left<<setfill(' ')<<setw(10)<<++count

cout<<i<<"       "<<j<<"       "<<k<<endl

}

cout<<endl<<"共有"<<count<<"种组合"<<endl<<endl

}

已调试完毕,共3003条!!!每输出40条按任一键继续...,保证结果万无一失,你可以查看下,源程序如下:

#include "stdio.h"

main()

{int i,j,k,m,n

long int count=0

for(i=1i<=15i++)

for(j=i+1j<=15j++)

for(k=j+1k<=15k++)

for(m=k+1m<=15m++)

for(n=m+1n<=15n++)

{if(iden(i,j,k,m,n))

printf("No %ld : %5d%5d%5d%5d%5d\n",++count,i,j,k,m,n)

if(count%45==0) getchar()

}

}

iden(int i,int j,int k,int m,int n)

{if(i==j||i==k||i==m||i==n||j==k||j==m||j==n||k==m||k==n||m==n)

return 0

else

return 1

}

再增加两个变量,修改下程序可以35选5,我机子是P4 cpu1.8G 内存是512M,跑了好久,估计要两个小时左右!!!没耐心等下去,有朋友想试下的我把程序弄下面来了!!

#include "stdio.h"

main()

{int i,j,k,m,n,q,w

long int count=0

for(i=1i<=35i++)

for(j=i+1j<=35j++)

for(k=j+1k<=35k++)

for(m=k+1m<=35m++)

for(n=m+1n<=35n++)

for(q=n+1q<=35q++)

for(w=q+1w<=35w++)

{if(iden(i,j,k,m,n,q,w))

printf("No %ld : %5d%5d%5d%5d%5d%5d%5d\n",++count,i,j,k,m,n,q,w)

if(count%45==0) getchar()

}

}

iden(int i,int j,int k,int m,int n,int q,int w)

{if(i==j||i==k||i==m||i==n||i==q||i==w||j==k||j==m||j==n||j==q||j==w||k==m||k==n||k==q||k==w||m==n||m==q||m==w||n==q||n==w||q==w)

return 0

else

return 1

}

void f()

{

int i1,i2,i3,i4,i5,i6,i7,i8

for (i1=1i1<32i1++)

for (i2=1i2<32i2++){if (i1==i2) continue

for (i3=1i3<32i3++){if (i1==i3) continueif (i2==i3) continue

for (i4=1i4<32i4++){if (i1==i4) continueif (i2==i4) continueif (i3==i4) continue

for (i5=1i5<32i5++){if (i1==i5) continueif (i2==i5) continueif (i3==i5) continueif (i4==i5) continue

for (i6=1i6<32i6++){if (i1==i6) continueif (i2==i6) continueif (i3==i6) continueif (i4==i6) continueif (i5==i6) continue

for (i7=1i7<32i7++){if (i1==i7) continueif (i2==i7) continueif (i3==i7) continueif (i4==i7) continueif (i5==i7) continueif (i6==i7) continue

for (i8=1i8<32i8++){if (i1==i8) continueif (i2==i8) continueif (i3==i8) continueif (i4==i8) continueif (i5==i8) continueif (i6==i8) continueif (i7==i8) continue

printf("%d,%d,%d,%d,%d,%d,%d,%d\n",i1,i2,i3,i4,i5,i6,i7,i8)

}}}}}}}

}

不过估计没有什么用，因为输出的组合数多达

31*30*29*28*27*26*25/(1*2*3*4*5*6*7*8)个

没有实际意义。

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