var n,m,i,j:longint; f:boolean; begin m:=0; randomize; readln(n); for i:=1 to n do begin f:=true; for j:=1 to 5 do if (random(7)+1)=j then begin f:=false;break;end; if f then inc(m); end; writeln(m/n); readln; end. 这叫频率估计概率
var j1,j2,j3,j4,j5,n,t:integer; begin readln; t:=0;n:=0; for j1:=2 to 7 do begin for j2:=1 to 7 do begin if j2<>j1 then for j3:=1 to 7 do begin if (j3<>j2)and(j3<>j1)then for j4:=1 to 7 do begin if(j4<>j1)and(j4<>j2)and(j4<>j3)then for j5:=1 to 7 do begin if(j5<>j1)and(j5<>j2)and(j5<>j3)and(j5<>j4)then begin inc(t); if(j1<>1)and(j2<>2)and(j3<>3)and(j4<>4)and(j5<>5)then inc(n); end; end; end; end; end; end; t:=(t div 6)*7; writeln(n,' ',t); readln; end. 这才是枚举吧......8楼居然使用了随机数组......真是在模拟瞎蒙啊...... 计算得出:n=1214,t=2520,也就是说,一共有2520种可能的答案,其中有1214种一分不得,目测将近一半啊。 也就是说,出现这种情况的概率为n/t*100%=48.17%,即607/1260