刘汝佳第二章习题

2-1 水仙花数，输出100~999中所有水仙花数

#include <iostream>
using namespace std;

int main (void)
{
    for (int i= 100; i <= 999; i++)
    {
       int a = i / 100;
       int b = ( i % 100 ) / 10;
       int c = i % 10;
       if ( i == a * a * a + b * b * b + c * c *c ){
           cout << i << ' ';
       }
    }
}

2-3 倒三角形 (提示：把图在纸上画出来，数一数几个星号，几个空，找规律

发现对于n层的倒三角形，第i行，有i-1个空格，有2 * n -2 * i + 1个星号即2(n-i)+1

#include <iostream>
using namespace std;

int main(void)
{
    int n;
    cin >> n；
    for ( int i = 1; i <= n; i++) {
        for ( int o = 1 ; o < i; o++ ) {
            cout << ' ';
        }
        
        for (int x = 1; x <= 2 * ( n -i ) + 1; x++) {
            cout << '*';
        }
        cout << endl;
    }
}

2-4 子序列的和

#include 

int main (void)
{
    int a, b;
    float sum=0;
    while(1){
        printf("Please input two numbers\n");
        scanf("%d %d", &a, &b);
        if (a==0 && b==0){
            return 0;
        }
        if (a >= b || b >= 10e6){
            printf("error\n");
            continue;
        } else {
            for (int i=a; i <= b; i++){
                sum+= (1.0/i) * (1.0/i);      
            }
            printf("%.5f\n", sum);
        }
    }
}

2-5 分数化小数

#include <stdio.h>

int main (void)
{
    while(1){
        int a, b, c;
        scanf("%d %d %d", &a, &b, &c);
        if (a == b == c == 0) break;
        printf("%.*f", c, (float) a/b); 
    }
}

2-6排列 (提示，用枚举法。因为三个数字各位的和为定值，乘积也为定值，从123扫描到987 / 3,发现满足上述条件的数字即可）

#include <iostream>
using namespace std;

int calc (int num, int &result_add, int &result_mul)
{
    int i = num / 100;
    int j = (num % 100) / 10;
    int k = num % 10;

    result_add = result_add + i + j + k;
    result_mul = result_mul * i * j * k;
}


int main (void)
{
    int i, j, k;
    int result_add;
    int result_mul;
    
    int sum = 0, pro = 1;
    for (int a = 1; a <= 9; a++) sum += a, pro *= a;

    for ( i = 123; i <= 987 / 3; i++ ){
        j = i * 2;
        k = i * 3;
        
        result_add = 0;
        result_mul = 1;
        calc(i, result_add, result_mul);
        calc(j, result_add, result_mul);
        calc(k, result_add, result_mul);
        if ( result_add == sum && result_mul == pro ){
            cout << i << ' ' << j << ' ' << k << endl;
        }
    }
}