C语言程序设计

Question

怎样存钱利最大 假设银行整存整取存款不同的月息利率分别为 0.63% _期限是一年 0.66%_期限是两年 另外一道; 歌星大奖 在歌星大奖赛中,有10个评委为参加的选手大分,分数为1-100分,选手最后得分为:去掉一个最高分和一个最底分后其余8个分数的平均直.请编写一个程序.
编写出个程序啊

Answer 1

1.存款n年(2的整数倍)
double lixi(double yuelilv,int qixian,int cunkuanshijian)
{
int m,n,i;
double ye=1;
n = cunkuanshijian/qixian;
m = 12 * qixian;
for(i = 0; i < n; i++)
{
ye += ye * m * yuelilv;
}

return ye-1;
}

void main()
{
int i;
for(i = 2; i < 20; i+=2)
{
printf("cun 1$ %d nian, li xi fen bie shi:%lf,%lf\n",i,lixi(0.0063,1,i),lixi(0.0066,2,i));
}
printf("press any key to exit:\n");
getch();
}

运行结果:
cun 1$ 2 nian, li xi fen bie shi:0.156915,0.158400
cun 1$ 4 nian, li xi fen bie shi:0.338453,0.341891
cun 1$ 6 nian, li xi fen bie shi:0.548477,0.554446
cun 1$ 8 nian, li xi fen bie shi:0.791457,0.800670
cun 1$ 10 nian, li xi fen bie shi:1.072564,1.085896
cun 1$ 12 nian, li xi fen bie shi:1.397781,1.416302
cun 1$ 14 nian, li xi fen bie shi:1.774030,1.799045
cun 1$ 16 nian, li xi fen bie shi:2.209318,2.242413
cun 1$ 18 nian, li xi fen bie shi:2.712909,2.756012
press any key to exit:

2.输入10个分数,输出平均分
void main()
{
int i,max,min,sum;
int s[10];

for(i = 0; i < 10; i++)
{
printf("Please input score ( %d ):\n", i + 1);
scanf("%d", &s[i]);
if(s[i] < 1 || s[i] > 100)
{
printf("not a score, input again!\n");
i--;
}
}

max = s[0];
min = s[0];
sum = 0;
for(i = 0; i < 10; i++)
{
if(max < s[i])
max = s[i];
if(min > s[i])
min = s[i];
sum += s[i];
}
sum = sum - min - max;

printf("average = %.2lf\n", sum/8.0);

getch();
}

Answer 2

第一题是不是缺少条件？当然是0.66%时利最大。
第二题先求10项的总和，在求出其中最大值，最小值，总和减去最大值和最小值的和除以8就是平均成绩。
下面是第二题代码：
------------
main()
{
int i,j,a[11],m,sum=0,max,min;
float ave;

for(i=0;i<10;i++)
{
printf("Input a number: 1-100:\n ");
scanf("%d",&j);
a[i]=j;
sum=sum+j;
if(i==9)
{

max = a[0];
min = a[0];
for(m= 0; m < 10; m++)
{
if(max < a[m])
max = a[m];
if(min > a[m])
min = a[m];
}
printf("The min number is :\n");
printf("%d\n",min);

printf("The max number is :\n");
printf("%d\n",max);
sum=sum-max-min;
ave=sum/8.0;
printf("The sum is :\n");
printf("%d\n",sum);
printf("The average is :\n");
printf("%.2lf\n",ave);

}
}
getch();

}