C語言助教題目代碼部分

題8000022012，求一個整數任意次方的最後三位數。

 

#include 
#include 

int main()
{
    int m,n;
    scanf("%d %d",&m,&n);
    int i;
    int sum = 1;
    for(i=0;i1000)
			sum = sum%1000;
		sum = sum * m;
	}
	printf("%d\n",sum%1000);
    return 0;
}

題8000022008 公式求π的近似值

 

 

#include 
#include 
#include
int main()
{
    double pi = 0;
    double i = 1.0;
    do
	{
		pi+= 1/i;
		if(i>0)
			i = -(i+2);
		else
			i = fabs(i) +2;
	}while(fabs(1/i)>1e-6);
	printf("%.6lf\n",4*pi);
    return 0;
}

題8000022007 求實數整數部分位數

 

 

#include 
#include 
#include
int main()
{
	double num;
	scanf("%lf",&num);
	int count = 0;
	int number ;
	number = (int)(fabs(num));
	while(number)
	{
		count++;
		number = number / 10;
	}
    printf("%d\n",count);
    return 0;
}

8000022002二分法求方程的根

 

 

#include 
#include
#include 

int  main()
{
	double x0,x1=0,x2=0,fx0,fx1,fx2;//x1,x2未端點，x0為中點,浮點型數據
	//判斷區間是否有解
	printf("請輸入端點x1,x2:\n");
	scanf("%lf %lf",&x1,&x2);
	printf("x1:%lf\n",x1);
	printf("x2:%lf\n",x2);
	fx1=2*x1*x1*x1-4*x1*x1+3*x1-6;
	fx2=2*x2*x2*x2-4*x2*x2+3*x2-6;
	if(fx1*fx2>0)
	{
		printf("have no anwser\n");
		exit(0);
	}
	//二分法求解
	do
	{
		x0=(x1+x2)/2;
		fx0=2*x0*x0*x0-4*x0*x0+3*x0-6;
		if (fx0*fx1<0)//在該區間
		{
			x2=x0;//調整端點位置
			fx2=fx0;
		}
		else
		{
			x1=x0;
			fx1=fx0;
		}
	} while (fabs(fx0)>1e-6);// 這裡精度太小可能跟文檔答案有偏差
	printf("方程的根%.2lf",x0);
	return 0;
}

題8000022001 迭代法求平方根的值

 

 

#include 
#include 
#include
int main()
{
	double num;
	scanf("%lf",&num);
	double x1,x2=1.0;
	do
	{
		x1 = x2;
		x2 = 0.5*(x1 + num/x1);

	}while(fabs(x1-x2)>1e-5);
	printf("%lf\n",x1);
    return 0;
}

題8000022000完數問題

 

 

#include 
#include 
#include

int isCompleteNum(int num)
{
	int i;
	int ans = num;
	for(i=1;i<= num/2;i++)
	{
		if(num%i==0)
			ans-= i;
	}
	if(ans == 0)
		return 1;
	else
		return 0;

}

void PrintFactor(int num)
{
	printf("%d=",num);
	int i,j=0;
	int factor[20];
	for(i=1;i<=num/2;i++)
	{
		if(num%i==0)
			factor[j++]=i;
	}
	for(i=0;i

題8000010008百分制成績轉換
 
 
#include 
#include 

int main()
{
	int grade;
	scanf("%d",&grade);
	printf("%c\n",grade);
	switch(grade/10)
	{
		case 10:
		case 9: printf("A\n");break;
		case 8:printf("B\n");break;
		case 7:printf("C\n");break;
		case 6:printf("D\n");break;
		default:printf("E\n");

	}
    return 0;
}



題8000010007 輸入日期，判斷是這一年的第幾天
 
 
#include 
#include 

int isLeapYear(int year)
{
	if((year%4==0&&year%100!=0) ||(year%400==0))
	{
		return 1;
	}
	else
	{
		return 0;
	}
}
int main()
{
	int year,month,day;
	scanf("%d-%d-%d",&year,&month,&day);

	int dayOfMonth[] = {31,28,31,30,31,30,31,31,30,31,30,31};

	int totalDays = 0,index;
	for(index=0;index2))
	{
		totalDays++;
	}

    printf("%d-%d-%d\n",year,month,day);
    printf("total days = %d",totalDays);
    return 0;
}