// 输入一个1-4位整数、如果满足 a=i*i*i*i 等、即输出结果、若不满足则加以说明



#include "stdio.h"
#include "windows.h"



void main()
{
    int a,i,j;
    printf ( "Please input a:" );
   
    scanf ( "%d" , &a );
    for ( i = 0 ; i < a ; i ++ )
    {
        if ( a == i * i )
        {    printf (" a = %d * %d ",i,i );
            Sleep(1000);
            printf("\n");
            for ( j = 0;j < a / 2;j ++ )
                if ( i == j * j )
                {
                    printf (" a == %d * %d * %d * %d ",j,j,j,j);
                    printf("\n");
                    Sleep(1000);
                }
        }
        else
        {
            printf("无相同数相乘为a ");
            Sleep(1000);
            printf("\n");}
    }



}




有办法只让最后一个if语句只判断一次么?我弄了很久都没头绪 。

这根本不需要什么循环！

#include"math.h"
#include"stdio.h"
void main()
{
    int a;double i,j;
    printf ( "Please input a:" );

    scanf ( "%d" , &a );
    i=sqrt(a);
    if ( a == (int)i*(int)i )
    {
        printf (" a = %d * %d ",(int)i,(int)i );
        printf("\n");
        j=sqrt(i);
        if ( i == (int)j *(int)j )
        {
            printf (" a = %d * %d * %d * %d ",(int)j,(int)j,(int)j,(int)j);
            printf("\n");
        }
    }
    else
    {
        printf("无相同数相乘为a ");
        printf("\n");
    }
}

#include<stdio.h>
#include<math.h>
main()
{
    int a,b,c;
    scanf("%d",&a);
    b=(int)sqrt(a);
    if(a==b*b)
    {
        c=(int)sqrt(b);
        if(b==c*c)
        {
            printf("%d=%d*%d*%d*%d\n",a,c,c,c,c);
        }
        else
            printf("%d=%d*%d\n",a,b,b);
    }
    else
       printf("无相同数相乘为a\n");
}

[ 本帖最后由 kinwyb 于 2010-4-10 10:38 编辑 ]