作业 2 分支、循环结构

      1.

#include<stdio.h>
int main(void)
{
 int numb1,numb2,sum,difference,produt,quotient,remainder;
 printf("Enter numb1,numb2:");
 scanf("%d%d",&numb1,&numb2);
 sum=difference=produt=quotient=remainder=0;
 sum=numb1+numb2;
 difference=numb1-numb2;
 produt=numb1*numb2;
 quotient=numb1/numb2;
 remainder=numb1%numb2;
 printf("sum=%d,difference=%d,produt=%d,quotient=%d,remainder=%d\n",sum,difference,produt,quotient,remainder);
 return 0;
}

 

         2.

#include<stdio.h>
int main(void)
{
 double Q,money;
 printf("Enter Q:");
 scanf("%lf",&Q);
 if(Q<=50){
  money=Q*0.53;
 }
 else {
  money=50*0.53+(Q-50)*0.58;
 }
 printf("money=%lf\n",money);
 return 0;
}

          3 .

#include<stdio.h>
int main(void)
{
 int n,x,i,y;
 printf("Enter n x:");
 scanf("%d%d",&n,&x);
 y=1;
 for(i=1;i<=n;i++){
  y=y*x;
  
 }
    printf("y=%d\n",y);
 return 0;
}

        4.

#include<stdio.h>
int main(void)
{
 int x,y,m,n,i;
 double h;
 printf("Enter m,n:");
 scanf("%d%d",&m,&n);
 if(m>n){x=m;y=n;}
 else{x=n;y=m;}
 h=0.00;
 for(i=y;i<=x;i++){
  h=h+i*i+1/i;
 }
 printf("h=%.2lf",h);
 return 0;
}

       5.

#include<stdio.h>
int main(void)
{
 int n,i,m;
 double y;
 printf("Enter n:");
 scanf("%d",&n);
 y=0;
 m=1;
 for(i=1;i<=n;i++){
  if(i%2==0){y=y-i/m;}
  else{y=y+i/m;}
  
  m=m+2;
 }
 printf("%.2lf\n",y);
 return 0;
}

          6.

#include<stdio.h>
#include<math.h>
int main(void)
{
 int year,month;
 double rate,loan,money;
 month=1;
 printf("Enter the rate and the loan:");
 scanf("%lf%lf",&rate,&loan);
 for(year=5;year<=30;year++){
  month=12*year;
  money=loan*rate*pow(1+rate,month)/(pow(1+rate,month)-1);
  printf("year=%dmoney=%.0lf\n",year,money);
 }
 return 0;
}

       7.

#include<math.h>
#include<stdio.h>
int main(void)
{
 int n,i;
 double y;
 printf("Enter n:");
 scanf("%d",&n);
 y=0;
 for(i=1;i<=n;i++){
  y=y+pow(2,i);
 }
 printf("%.2lf\n",y);
 return 0;
}

     8.

#include<stdio.h>
double fact(int n);
int main(void)
{
 int i,n;
 double e;
 printf("Enter n:");
 scanf("%d",&n);
 e=0;
 for(i=0;i<=n;i++){
  e=e+fact(i);
 }
 printf("%.2lf\n",e);
 return 0;
}
double fact(int n)
{
 int i;
 double y;
 y=1;
 for(i=1;i<=n;i++){
  y=y*i;
 }
 return y;
}