#include <iostream>
#include<stdio.h>

using namespace std;

int main()
{
    int num=100;

    for(;num<=999;num++)
    {
        int n1=num%10;
        int n3=num/100;
        int n2=(num/10)%10;

        if(num==n1*n1*n1+n2*n2*n2+n3*n3*n3)
            printf("%d ",num);
    }

    printf("
 
\n");
    return 0;
}

習題2-1

#include<iostream>
#include<stdio.h>


int main()
{
    int a,b,c;
    

                //以下循環在輸入小於兩個數時會自動跳過制表空格和回車
                //在輸入大於三個數時會把多余的數留在緩沖區中,在鍵入回車時結束
                //在輸入三個數時會在鍵入回車是結束,標準輸入中
                //另外在輸入不符合格式化字符串中的格式時也會退出

    while(scanf("
 
%d%d%d",&a,&b,&c)==3)    
    {
        int i;

        for( i=10;i<=100;i++)
        {
            if(i%3==a && i%5==b && i%7==c)
                {
                    printf("%d\n",i);
                    break;
                }
        }

        if(i>100)
            printf(
 
"No answer\n");
    }
}

習題2-2

#include<iostream>
#include<stdio.h>

void printfsy(int n);
void printfsp(int n);

int main()
{
    int n;
    scanf("%d",&n);

    for(int i=n,j=0;i>=1;i--,j++)
    {
        int num=i*2-1;
        printfsp(j);
        printfsy(num);
        printf("\n");
    }
}

void printfsy(int n)
{
    for(int i=1;i<=n;i++)
        printf("#");
}

void printfsp(int n)
{
    for(int i=1;i<=n;i++)
        printf(" ");
}

2-3

#include<iostream>
#include <stdio.h>

int main()
{
    int n,m;

    while(scanf("%d%d",&n,&m)==2 && (n|m)!=0)
    {
        double s=0.0;

        //n*n 對於int會溢出,對long也會溢出,所以要用long long
        //另外The compiler looks at the literal value 655360*655360 
        //without considering the variable that you‘re assigning it
        // to/initializing it with. You‘ve written it as an int typed   literal,
        //  and it won‘t fit in an int.Use 655360LL*655360LL

        for(;n<=m;n++)
        {
            //long long mul=(long long)n*(long long)n;

            s+=1.0/n/n;
        }

        printf("%.5f\n",s);
    }

    return 0;
}

2-4所謂陷阱就是溢出註意註釋部分的內容

#include<iostream>
#include <stdio.h>

int main()
{
    for(int i=123;i<=987;i++)
    {
        if(i*3<=987)
        {
            int sn=i;
            int mn=i*2;
            int ln=i*3;

            int sn_s=sn%10;
            int sn_m=(sn/10)%10;
            int sn_l=sn/100;

            int mn_s=mn%10;
            int mn_m=(mn/10)%10;
            int mn_l=mn/100;

            int ln_s=ln%10;
            int ln_m=(ln/10)%10;
            int ln_l=ln/100;

            if(ln_l*ln_m*ln_s*mn_l*mn_m*mn_s*sn_m*sn_l*sn_s==362880
                && ln_l+ln_m+ln_s+mn_l+mn_m+mn_s+sn_m+sn_l+sn_s==45)
                printf("%d %d %d\n",sn,mn,ln);
        }

    }

    return 0;
}

2-5

第二章習題