《算法竞赛入门经典 第二版》习题——Chapter 2

习题2-1 水仙花数(daffodil)

#include<iostream>#include<cstdio>using namespace std;int main(){int a, b, c;for (int num = 100; num < 1000; num++){a = num / 100;b = num / 10 % 10;c = num % 10;if (num == a*a*a + b*b*b + c*c*c)cout << num << " "；}cout << endl;}

#include<iostream>#include<cstdio>using namespace std;int main(){for (int num = 100; num < 1000; num++){int temp = num;int s = 0, a;while (temp){a = temp % 10;s += a*a*a;temp /= 10;}if (num == s)cout << num << " ";}cout << endl;}

习题2-2 韩信点兵(hanxin)

#include<iostream>#include<cstdio>using namespace std;int main(){int a, b, c;int count = 0;while (cin >> a >> b >> c){++count;int i;for (i = 10; i <= 100; i++){if (i % 3 == a && i % 5 == b && i % 7 == c){printf("Case %d: %d\n", count, i);break;}}if (i == 101){printf("Case %d: No answer\n", count);}}}

习题2-3 倒三角形(triangle)

#include<iostream>#include<cstdio>using namespace std;int main(){int n,b=0;cin >> n;for (; n > 0; n--,b++){for (int i = 0; i < b; i++)cout << " ";for (int j = 2*n-1; j >0; j--){cout << "#";}cout << endl;}return 0;}

习题2-4子序列的和(subsequence)

分析：本题陷阱在于n比较大时，n*n会溢出，可以用1/n/n替代1/n^2，或者将n*n强制转换为double

#include<iostream>#include<cstdio>using namespace std;int main(){int  n, m;int kase = 0;while (cin >> n >> m && m!=0 || n!=0){++kase;double s = 0;for (long long  i = n; i <= m; i++){s += 1 / double(i*i);}printf("Case %d: %.5f\n", kase, s);}return 0;}

习题2-5 分数化小数(decimal)

分析：printf("%*.*lf", x, y, z) 第一个*对应x，第二个*对应y，lf对应z （高级特性）

注：该方案不完善，有 bug~

#include<iostream>#include<cstdio>using namespace std;int main(){int  a, b;int c = 0,kase=0;while (cin >> a >> b >> c && a!=0 || b!=0 || c!=0){++kase;double f =1.0 * a /b;printf("Case %d: %.*lf",kase, c, f);}return 0;}

习题2-6 排列(permulation)

分析：思路其实就是穷举，首先可以判断出最小的数在123~329之间。利用一个数组a[1]~a[9]，值为0表示没出现过，为1则表示出现，注意最后要清零！

#include<iostream>#include<cstdio>using namespace std;int main(){int a[10] = { 0 };int b;for (int i = 123; i <= 329; i++){for (int j = 1; j <= 3; j++){int temp = i*j;while (temp){b = temp % 10;a[b] = 1;temp /= 10;}}b = 0;for (int k = 1; k < 10; k++)b += a[k];if (b == 9)printf("%d %d %d\n", i, i * 2, i * 3);for (int k = 0; k < 10; k++)a[k] = 0;}return 0;}