第4次作业

3.1源码：#include <stdio.h>

int main()

{

       int i,sum=0;

       i=1;

       while(i<=100)

       {

       sum+=i;

       i++;      

       }

       printf("sum=%d\n",sum);

       return 0;

}#include <stdio.h>

int main()

{

       int i,sum=0;

              do{

              sum+=i;

              i++;

       }

       while(i<=100);

       printf("sum=%d\n",sum);

       return 0;

}#include <stdio.h>

int main()

{

       int i,sum=0;

       for(i=1;i<=100;i++)

       sum+=i;

       printf("sum=%d\n",sum);

       return 0;

}

运行结果抓图

3.2源码#include <stdio.h>

int main()

{

       int i,n,nj,sum=1;

       printf("putin n:");

       scanf("n=%f",&n);

       while(i<=n)

       {

       sum+=i;

       i++;

       nj=sum*i;

       }

       if(i=n)

       printf("n!=%f\n",nj);

       return 0;

}

3.3程序源码#include<stdio.h>

int main(){

      int k,h;

      int sum=0,sum1=0,sum2=0;

      for(k=1;k<=100;k++)

         sum+=k;

 printf("sum=%d\n",sum);

 for(k=1;k<=50;k++)

 sum1=sum1+k*k;

 printf("sum1=%d\n",sum1);

 for(k=1;k<=10;k++)

 sum2+=1/k;

  printf("sum2=%d\n",sum2);

h=sum+sum1+sum2;

printf("结果为：%d\n",h);

return 0;

3.4程序源码#include <stdio.h>

int main()

{

       char c;

       int zm=0,sz=0,kg=0,qt=0;

       while((c=getchar())!=‘\n‘)

       {

              if((c>=‘a‘&& c<=‘z‘)||(c>=‘A‘&& c<=‘Z‘))

              zm++; else if(c==‘ ‘)

              kg++;

              else if(c>=‘0‘&&c<=‘9‘)

              sz++;

              else

              qt++;

}

       printf("the number of letter:%d\n",zm);

printf("空格个数;%d\n",kg);

       printf("数字个数;%d\n",sz);

       printf("其他个数;%d\n",qt);

       return 0;

}

3.5程序源码#include<stdio.h>

int   main(){

       int m,n,t,i;

       printf("请输入两个整数：");

       scanf("%d%d",&m,&n);

       if(n>m){

              t=m;

              m=n;

              n=t;

       }

       for(i=n;i>1;i--)

       if(m%i==0&&n%i==0)

       break;

       printf("m=%d,n=%d",m,n);

       if(i!=0)

       printf(",最大公因数为：%d",i);

       t=(m*n)/i;

       printf("最小公倍数为: %d",t);

       return 0;

}

3.6程序源码#include <stdio.h>

int main(void)

{

    int a,b,c;

    for(a=1;a<=9;a++)

    for(b=0;b<=9;b++)

    for(c=0;c<=9;c++)

    if(100*a+10*b+c==a*a*a+b*b*b+c*c*c)

    printf("%d%d%d\n",a,b,c);

    return 0;

}

实验心得

结构大体清楚，但题目里的算法和意思不太懂。