汉诺塔问题的求解程序.doc

1. 基于汉诺塔问题的求解程序，分别使用全局变量和局部变量来计算当塔中有n个盘子时需要进行多少步的盘子移动次数（例如当塔中有5个盘子时，需要移动31次）。提示：当使用局部变量时，函数的原型可以为 long hanoi(int ,char,char,char),该函数的功能是完成hanoi塔的移动并返回其执行步数。方法一： 使用局部静态变量#includelong hanoi(int n,char one, char two, char three) /*该函数计算移动盘的步骤并返回移动步数 */ static long int count=0; /* 定义静态局部变量count，用以计算总共移动他多少次 */ if(n=1) printf(move %c- %cn,one, three); count+; else hanoi(n-1,one,three,two); printf(move %c- %cn,one, three); count+; hanoi(n-1,two,one,three); return(count);void main() int m=0; long int steps; /*定义长整型变量step用来装载hanoi函数的返回值*/ printf(please input the number of diskes:); scanf(%d,&m); printf(the steps are following:n); steps=hanoi(m,A,B,C); printf(They need %ld steps to moven,steps);方法二： 使用一般局部变量#includelong hanoi(int n,char one, char two, char three) /*该函数计算移动盘的步骤并返回移动步数 */ long int count1=0, count2=0; /* 定义局部变量count1，count2 */ if(n=1) printf(move %c- %cn,one, three); else count1=hanoi(n-1,one,three,two); printf(move %c- %cn,one, three); count2=hanoi(n-1,two,one,three); return(count1+count2+1);void main() int m=0; long int steps; /*定义长整型变量step用来装载hanoi函数的返回值*/ printf(please input the number of diskes:); scanf(%d,&m); printf(the steps are following:n); steps=hanoi(m,A,B,C); printf(They need %ld steps to moven,steps);方法三： 使用全局变量#includelong count=0; /*定义全局变量来统计移动步数 */void hanoi(int n,char one, char two, char three) /* 该函数计算移动盘的步骤*/ if(n=1) printf(move %c- %cn,one, three); count+; else hanoi(n-1,one,three,two); printf(move %c- %cn,one, three); count+; hanoi(n-1,two,one,three); void main() int m=0; printf(please input the number of diskes:); scanf(%d,&m); printf(the moving