求解C语言程序题
\u6c42\u89e3\u4e00\u4e2aC\u8bed\u8a00\u7a0b\u5e8f\u9898\u3002\u63d0\u4f9b\u4f60\u4e00\u4e2a\u601d\u8def\uff0c\u5c06
1,2,3,4
5,6,7,8
9,10,11,12
13,14,15,16
\u3002\u3002\u3002\u3002\u3002\u3002
\u4f5c\u4e3a\u6570\u7ec4\uff0c\u6309\u7167\u7f16\u6392\u597d\u7684\u65b9\u5f0f\u6253\u5370\u51fa\u6765\u3002
N\u4e2a\u7684\u8bdd\uff0c\u6570\u7ec4\u4e3a4\u4e2a\u957f\u5ea6n-1 \u7684\u6570\u7ec4\uff0c4\u4e2a\u957f\u5ea6n-3\u7684\u6570\u7ec4\u3002\u7c7b\u63a8
\u6216\u8005\u4f60\u505a\u4e00\u4e2a\u4e8c\u7ef4\u6570\u7ec4\uff0c\u6309\u7167\u987a\u5e8f\u751f\u6210\u6570\u503c\uff0c\u5728\u6253\u5370\u51fa\u6765
const int MAX=10; //
int d,i,j,m,N;
int a[][] = new a[MAX][MAX]; ;
Consloe.Write('Enter N:');
Consloe.Readline(N);
if ((N>=1) && (N<=max))
{
i=1;
j=1;
m=1;
d=1;
for(m=1;m<N*N;m++)
{
A[i,j]=m;
switch(d){
case 1: {i=i+1; if (j==1) d=2 else d=4; break};
case 2: {i=i-1;j=j+1; if (j==N) d=1 else if (i==1) then d=3;break;}
case 3: {j=j+1; if (i==N) d=2 else d=4;break};
case 4: {i=i+1;j=j-1;if (i==N) d==3 else if (j==1) d=1;break};
default: break;
}
}
Consloe.Writeline('OUTPUT:');
//for\u5faa\u73af\u8f93\u51fa\u6570\u7ec4
}
else Consloe.Writeline('Input N error!');
\u6709\u51e0\u5e74\u6ca1\u7528C#\u4e86\uff0c\u4f60\u81ea\u5df1\u6539\u4e00\u4e0b\u3002\u5c31\u662f\u5206\u62104\u4e2a\u65b9\u5411\u4e0d\u65ad\u5faa\u73af\u8f93\u51fa\u5230\u6570\u7ec4
\u8fd9\u9898\u6709\u70b9\u65e0\u5398\u5934\u554a\uff0c
\u9996\u5148\u770b\u4ee5\u4e00\u4e2a\u5faa\u73af \u5c31\u662f i=0\u5f00\u59cb\uff0c\u627e\u5230\u548c*a == b[i]
a\u662f\u201cyou\u201d\u7684\u9996\u5730\u5740\uff0c\u90a3\u4e48*a = 'y'\u7684
\u90a3\u4e48b[7] = \u2018y\u2019
\u4e5f\u5c31\u662f\u8bf4 if(*a == b[i])\u6210\u7acb\u65f6\uff0ci=7,b[i] = 'y'
\u90a3\u4e48\u6211\u4eec\u518d\u770b if\u91cc\u9762\u7684\u5faa\u73affor(j = i; *p!='\0';j++\uff09
{
if(*p != b[j]) break;
p++;
}
\u56e0\u4e3ap=a\u7684\uff0c\u90a3\u4e48*p=\u2018y\u2019\uff0cb[j] = 'y' p++\u4e4b\u540e *p =\u2018o\u2019 ,\u800cj++\u4e4b\u540e\uff0cb[j] ='o',\u90a3\u4e48\u4f1a\u4e00\u76f4\u8fd0\u884c\u5230\u201cyou\u201d\u7684\u7ed3\u5c3e \uff0c\u6700\u540e *p = '\0'
\u7136\u540e if(*p == '\0')\u6210\u7acb \u90a3\u4e48\u8df3\u51fa\u5927\u7684\u5faa\u73af
\u6700\u540e\u56e0\u4e3ai=7 \u90a3\u4e48printf("%s",&b[i]); \u5c31\u662f\u4eceb[7]\u5f00\u59cb\u5f80\u540e\u8f93\u51fa \u4e5f\u5c31\u662f you to china!
main()
{
int
p,s,i,a=0;/*p是输入的数字，s是p和1~10的乘积，a是用来控制输出换行的*/
printf("请输入一个数p:");
scanf("%d",&p);
for(i=1;i<=10;i++)/*这是一个for循环，就是i初始等于1，然后执行花括号内的语句，没执行一次自加1，结束条件是知道i=10*/
{
s=1;
s=s*p;
printf("%d*%d=%d",p,i,s);
a++;
if(a%2==0)
printf("\n");/*当a变为2的倍数的时候输出一次换行*/
}
}
#include <stdio.h>
int main()
{
int r = 0;
float a = 0, l = 0;
scanf("%d", &r);
a = r * r * 3.14;
l = r * 2 * 3.14;
printf("面积:%f, 周长:%f", a, l);
return 0;
}
#include <stdio.h>
void main(){
double sum,pri;
float y;
printf("请输入存款数目:");
scanf("%lf",&pri);
sum=pri；
printf("请输入年利率:");
scanf("%f",&y);
sum=sum*(1+f);
printf("一年过后你的资金总额为:
%lf\n",sum);
}
1.编写程序输入圆的半径，求圆的面积和周长
#include<stdio.h>
#define pai 3.14
void main()
{
float r,area,circ;
printf("Please input the radius:\n");
scanf("%f",&r);
area = pai*r*r;
circ = 2*pai*r;
printf("The area is %.2f,the circumference is %.2f\n",area,circ);
return;
}
2.编写程序输入年利率（假设3%），存款数为S（假设10000），计算一年后的本息合计并输出
#include<stdio.h>
#define pai 3.14
void main()
{
float rate = 0.03;//利率
int S = 10000;//本金
int year = 1;//年
float Sum;//本息合计
Sum = S + S * rate * year;
printf("Sum is %.2f\n",Sum);
return;
}
这些都是很基础的C语言语句，希望你还是能好好听课！
1.
#include<stdio.h>
int main()
{
int r;
float s,l;
printf("请输入圆的半径:")
scanf("%d",&r);
printf("\n");
l=3.1416*r*r;
s=3.1416*r*2;
printf("圆的面积为:%f\n",s);
printf("圆的周长为：%f\n",l);
return 0;
}
2.
#include<stdio.h>
#define S 10000
int main()
{
int i;
i=S*(1+0.03);
printf("本息为：%d",i);
return 0;
}
1
#include<stdio.h>
main()
{float s,l,r,pi;
printf("please input r\n");
scanf("%f",&r);
pi=3.14;
s=pi*pi*r;
l=2*pi*r;
printf("area=%f,long=%f\n",s,l);
}
2
#include<stdio.h>
main()
{ float p,s,total;
printf("please input p and s\n");
scanf("%f,%f",&p,&s);
total=s+s*p;
printf("total=%f\n",total);
}
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