1、求阶乘
Console.WriteLine("输入一个数"); int n = Convert.ToInt32(Console.ReadLine()); int s = 1; for (int i = 1; i <= n; i++) { s = s * i; } Console.WriteLine("结果:" + s);
例:求5!
2、求阶乘的和
Console.WriteLine("输入一个数:"); int n = Convert.ToInt32(Console.ReadLine()); int a = 0; for (int i = 1; i <= n; i++) { int s = 1; for (int j = 1; j <= i; j++) { s = s * j; } a = a + s; } Console.WriteLine("结果:" + a);
例:5!+4!+3!+2!+1!=
3、找出100以内质数,并求和
int sum = 0; for (int i = 2; i <= 100; i++) { int a = 0; for (int j = 1; j <= i; j++) { if (i % j == 0) { a++; } } if (a == 2) { sum = sum + i; Console.Write(i + " "); } } Console.Write("总和:" + sum);
4、100元购物券买香皂(2元)、牙刷(5元)、洗发水(15元),每样至少买一个,正好花光,求所有可能
int n = 1; for (int i = 1; 2 * i < 100; i++) { for (int j = 1; 5 * j < 100; j++) { for (int k = 1; 15 * k < 100; k++) { if (2 * i + 5 * j + 15 * k == 100) { Console.WriteLine(n); Console.WriteLine("香皂:" + i); Console.WriteLine("牙刷:" + j); Console.WriteLine("洗发水:" + k); n++; } } } }
5、公鸡2文,母鸡1文,小鸡半文,每种至少买一只,100文钱买100只鸡,求所有可能
int n = 1; for (int i = 1; 2 * i < 100; i++) { for (int j = 1; j < 100; j++) { for (int k = 1; 0.5 * k < 100; k++) { if (2 * i + j + 0.5 * k == 100 && i + j + k == 100) { Console.WriteLine(n); Console.WriteLine("公鸡{0}只,母鸡{1}只,小鸡{2}只", i, j, k); //"{0}",占位符,必须从0开始 n++; } } } }