很有意思的一个题目，呵呵

应该是一次概率试验问题吧？？

应该是组合数学里边的问题

不是高手，在室友的数学指导下，两天才搞定，还不知道对不对~~

/*若n+1个数有公约数，必败；成功的充要条件是n+1个数互质*/
/*证明充分性：若只有两个数，互质，辗转相减可得1（反证）；
当有n+1个数互质时，任意n个数可凑出一个数与第n+1个数互质（否则就存在公约数k，使得不互质）*/
/*解题思路：m**n-不互质的方案数*/
/*不互质方案数：找出M所有唯一质因数，不互质方案里的所有数一定有共同的质公约数*/
/*小范围试验，归纳公式，求补集不重不漏*/
/*K个质数多种组合相乘的办法实现，比较土*/
/*10,000,000以内的数，唯一质因数个数不超过8个*/
%macro tiaozao(m,n);
data a1;
m=&m;
do i=2 to m;
if mod(m,i)=0 then do
m=m/i;
output;
i=i-1;
end;
end;
run;
data a2;
set a1 end=last ;
if i>lag(i);
if i>lag(i) then x+1;
if last then call symput("k",x);
drop m x;
run;
%macro repli;
%do p=1 %to &k;
data c&p(rename=(i=i&p));
set a2;
if _n_<&p then i=1;
run;
%end;
%do q=&k+1 %to 8;
data c&q;
i&q=1;
%end;
run;
%mend;
%repli;
proc sql;
create table base1 as select * from c1,c2,c3,c4,c5,c6,c7,c8;
quit;
data base2;
set base1;
array i(8);
do r=1 to 8;
do s=r+1 to 8;
if (i(r)=i(s) and i(r)>1) then delete;
end;
end;
drop r s;
product=i1*i2*i3*i4*i5*i6*i7*i8;
run;
proc sort data=base2 out=base3;
by product i1 i2 i3 i4 i5 i6 i7 i8;
run;
data base4;
set base3;
by product i1 i2 i3 i4 i5 i6 i7 i8;
if first.product;
run;
data base5;
set base4;
count=0;
if i1=1 then count+1;
if i2=1 then count+1;
if i3=1 then count+1;
if i4=1 then count+1;
if i5=1 then count+1;
if i6=1 then count+1;
if i7=1 then count+1;
if i8=1 then count+1;
paraNo=8-count;
drop count ;
run;
proc sort data=base5 out=base6;
by paraNo product;
run;
data base7;
set base6 ;
m=&m;
n=&n;
total=m**n;
factor=(-1)**(paraNo-1)*(m/product)**n;
sum+factor;
run;
data final;
set base7 nobs=nobs;;
final=total-sum;
keep final;
if _n_=nobs;
run;
%mend tiaozao;
%tiaozao(4,2);