作业 2 分支、循环结构

/*两个整数的四则运算*/

#include<stdio.h> int main(void) { int num1,num2,h,c,j,y; /*h和,c差,j积,s商,y余数*/
double s; printf("Enter num1 and num2:"); scanf("%d%d",&num1,&num2); h=num1+num2; c=num1-num2; j=num1*num2; s=num1/num2; y=num1%num2; printf("h=%d,c=%d,j=%d,s=%Lf,y=%d\n",h,c,j,s,y); return 0; }

两个数都是整型时,除法结果也是整型

 

/*计算x的n次方*/

#include<stdio.h>

int main()

{

    int n,i;

    double x,y;

    y=1;

    printf("Enter n and x:");

    scanf("%d%Lf",&n,&x);

    for(i=1;i<=n;i++)

    y=y*x;

    printf("y=%.2f\n",y);

    return 0;

}

  

/*求i^2+1/i*,i=m~n,求和/

#include<stdio.h>

int main()

{

    int m,n,i;

    double y;

    y=0;

    printf("Enter m and n:");

    scanf("%d%d",&m,&n);

    for(i=m;i<=n;i++)

    y=y+(i*i+1.0/i);     //转换类型
    printf("y=%.2f\n",y);

    return 0;

     

}

  

/*计算1-2/3+3/5-4/7+5/9-6/11+^的前n项和*/

#include<stdio.h> int main() { int n,flag; double item,sum,i; printf("Enter n:"); scanf("%d",&n); flag=1; /*第i项的符号*/ sum=0; for(i=1.0;i<=n;i++){ item=flag*i/(2*i-1); /*第n项的值*/ sum=sum+item; flag=-flag; } printf("sum=%f\n",sum); return 0; }

  

/*打印还款年限-月还款额表*/

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

     

/*计算2^i,i=1~n,输入n并求和*/

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

  

posted @ 2013-10-19 20:54  墨墨萧萧  阅读(182)  评论(0)    收藏  举报