用VB程序设计一个歌手大赛的评分程序

’3个label，2个command

Private Sub Command1_Click()

Dim s, i, j, arr(1 To 10), t, sum

For i = 1 To 10

s = InputBox("输入评分（0-10分）：", "第" &i &"个分数")

If IsNumeric(s) Then

s = Val(s)

If s <0 Or s >10 Then

i = i - 1

Else

arr(i) = s

End If

Else

i = i - 1

End If

Next

For i = 1 To 9

For j = i + 1 To 10

If arr(i) >arr(j) Then

t = arr(i): arr(i) = arr(j): arr(j) = t

End If

Next

Next

For i = 2 To 9

sum = sum + arr(i)

Next

Label1.Caption = "最高分亩亏侍为：" &arr(10)

Label2.Caption = "最低分为：" &arr(1)

Label3.Caption = "选手最后得分："迅吵 &Round(sum / 8, 1)

End Sub

Private Sub Command2_Click()

Unload Me

End Sub

Private Sub Form_Load()

Command1.Caption = "评分"

Command2.Caption = "结束"空铅

With Label1

.Caption = ""

.AutoSize = True

End With

With Label2

.Caption = ""

.AutoSize = True

End With

With Label3

.Caption = ""

.AutoSize = True

End With

End Sub

#include<stdio.h>

#include<conio.h>

#define _for(i,a,b) for(int i=ai<bi++)

using namespace std

int scores[10]

double average,maxn,minn=0x7f,imax,imin

int main(){

system("cls")

_for(i,0,10){

printf("Please input a score:")

scanf("%d",&scores[i])

average+=scores[i]

if(maxn<scores[i]){

maxn=scores[i]

imax=i

}

if(minn>scores[i]){

minn=scores[i]

imin=i

}

printf("\n")

}

average-=maxn+minn

average/=8.0

printf("\锋谨nThe average: %.6lf\n",average)

printf("The max: %.6lf\n"银闭基,maxn)

printf("The min: %.6lf\n",minn)

_for(i,0,10){

if(scores[i]>态如average){

printf("%d,%.6lf\n",i+1,double(maxn))

}

}

printf("max: %d,%.6lf\n",imax+1,maxn)

printf("min: %d,%.6lf\n\n",imin+1,minn)

printf("Next singer? <y or Y>: ")

char k=getch()

if(k=='y' || k=='Y')main()

return 0

}

看看这个能不能满足你的要求：

#include"stdio.h"

void main()

{float a[12][7],b[12],t

int i,j,k

float avr(float (*x)[7])

void arry(float px[])

printf("评委请打分：\n")/*按选手编号打行型分，如，一号选手得分，分别有7个评委打分，然后是二号选手；*/

for(i=0i<12i++)

{ for(j=0j<7j++)

scanf("%d",&a[i][j])

b[i]=avr(a+i)/*计算每个选手的平均成绩*/

printf("%d号选手最终得分：%d\n",i+1,b[i])

}

for(i=11i>=0i--)

{k=i

for(j=ij<12j++)

{if(b[j]<b[k])

k=j

}

if(b[k]!=b[i])

{t=b[i]b[i]=b[k]b[k]=t}

}

for(i=0i<12i++)

{switch(i)

{case 0: printf("一等奖的获得者是：%d\n",i+1)break

case 1: printf("二等奖的获得者是：%d号选手\n",i+1)break

case 2: printf("二等奖的获得者是：%d号选手\n",i+1)break

case 3: printf("三等奖的获得者是：%d号选手\n",i+1)break

case 4: printf("三等奖的获得乎带握者是：岁庆%d号选手\n",i+1)break

case 5: printf("三等奖的获得者是：%d号选手\n",i+1)break

case 6: printf("三等奖的获得者是：%d号选手\n",i+1)break

default: break

}

}

}

float avr(float (*x)[7])

{int i,m,n,t,j,k

float sum=0

for(i=0i<7i++)

{m=(*x)[0]

n=(*x)[0]

if((*x)[i]>m)

j=i

if((*x)[i]<n)

k=i

}

{t=(*x)[6](*x)[6]=(*x)[j](*x)[j]=t}

{t=(*x)[5](*x)[5]=(*x)[k](*x)[k]=t}

for(i=0i<5i++)

sum+=(*x)[i]

return sum/5

}

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