var a,total,f,sum:array [1..50] of integer; d:real; i,j,max,k,n,c,b,z,h,l,u:integer; s:array [1..50] of string;begin writeln('how many numbers?'); readln(n); for i:= 1 to n do begin writeln('input data'); readln(a[i]); end; for i:= 1 to n-1 do for j:= i+1 to n do begin if a[i]
} writeln('zhongshu is ',total[1],' ','times is ',total[2]+1); readln;end.

N多的缺陷,你试一下运行就知道了~~~求的是众数和中位数即例如输入: 1 1 1 2 2 2 4 5 6 就输出: 1 2 就是说,在一堆数,出现次数最高的数就是众数

建议使用冒泡排序

上边已经把这堆数排好序的了,好象是用冒泡法,但是上边排好序,与下边没关啊~~~最后请各位大哥大姐帮帮我~~不怕你错,只怕你不发

中位数没问题，问题在众数，你是希望total[1]做寄存众数大小的变量吧，这样的话问题就来了；试试这样：{------------------------------zhongshu-----------------------} total[2]:=1;total[4]:=0; for i:=2 to n do if a[i]=a[i-1] then total[2]:=total[2]+1 else begin {if total[2]=total[4] then total[5]:=total[4];} if total[2]>total[4] then begin total[3]:=total[1];total[4]:=total[2];end; total[1]:=a[i];total[2]:=1; end; total[1]:=total[3]; total[2]:=total[4]-1;其中total[5]是为了验证有无众数的如果输出不改，这个值没用！可以简化，有问题+我QQ：372161965或517168603

program MedianAndMode; //中位数和众数var a, x : array [0..1000] of integer; n, i, j, p, q, y : integer;begin write('How many numbers? '); readln(n); for i:=1 to n do begin write('Number ', i, ': '); readln(a[0]); j := i - 1; while a[j] > a[0] do begin a[j+1] := a[j]; j := j - 1 end; a[j+1] := a[0] end; if n mod 2 = 1 then writeln('Median is ', a[n div 2 + 1]) else writeln('Median is ', (a[n div 2]+a[n div 2 + 1])/2:0:1); p := a[1]; q := 1; y := 0; a[n+1] := abs(a[n]) - 1; for i:=2 to n+1 do if a[i] = p then q := q + 1 else begin if q > y then begin y := q; x[0] := 1; x[1] := p end else if q = y then begin x[0] := x[0] + 1; x[x[0]] := p end; p := a[i]; q := 1 end; write('Mode is/are ', x[1]); if x[0] > 1 then for i:=2 to x[0] do write(', ', x[i]); writeln('; The frequency is ', y)end.How many numbers? 13Number 1: 2Number 2: 3Number 3: 7Number 4: 7Number 5: 4Number 6: 3Number 7: 1Number 8: 6Number 9: 3Number 10: 6Number 11: 6Number 12: 7Number 13: 1Median is 4Mode is/are 3, 6, 7; The frequency is 3

嗯,谢谢大家...