求每只队伍都回答出题目，且至少有一只队伍回答出n道题的概率

存在性问题我们可以转化为任意性问题

用P(每支队伍都回答出题目)-P(每只队伍回答的题目数小于n)

然后我们可以递推求解

1 var f:array[0..1010,0..35,0..35] of double; 2     p:array[0..1010,0..35] of double; 3     k,i,j,n,m,t:longint; 4     s,ans,tmp:double; 5  6 begin 7   while not eof do 8   begin 9     readln(n,t,m);10     if (n=0) and (t=0) and (m=0) then break;11     for i:=1 to t do12     begin13       for j:=1 to n do14         read(p[i,j]);15       readln;16     end;17     fillchar(f,sizeof(f),0);18     for i:=1 to t do19     begin20       f[i,0,0]:=1;21       for j:=1 to n do22       begin23         f[i,j,0]:=f[i,j-1,0]*(1-p[i,j]);24         for k:=1 to j do25           f[i,j,k]:=f[i,j-1,k-1]*p[i,j]+f[i,j-1,k]*(1-p[i,j]); //f[i,j,k]表示第i支队前j道题答出k题的概率26       end;27     end;28     ans:=1;29     for i:=1 to t do30       ans:=ans*(1-f[i,n,0]);31     tmp:=1;32     for i:=1 to t do33     begin34       s:=0;35       for j:=1 to m-1 do36         s:=s+f[i,n,j];  //答出题目出数