1-100用c语言的递归法求和

1、首先本题的思路已经给出，要求用递归算法。

2、程序源代码如下：

#include<stdio.h>

int sum(int n)

int main()

{

int i=100

printf("1~%d的和为：\n",i)

printf("%d\n",sum(100))

return 0

}

//下面是一个递归求和函数

int sum(int n)

{

if (n==1)

return 1

else

return n+sum(n-1)

}

3、程序的输出结果如下：

递归算法是我前些天写的，非递归是刚才找的，里面含递归和非递归。\x0d\x0a递归算法：\x0d\x0a#include \x0d\x0a//递归求汉诺塔问题\x0d\x0avoid hanoi(int n, char A, char B, char C, int *time)\x0d\x0a{\x0d\x0aif (n>=1)\x0d\x0a{\x0d\x0a hanoi(n-1, A, C, B, time)\x0d\x0a move(A, C)\x0d\x0a (*time)++\x0d\x0a hanoi(n-1, B, A, C, time)\x0d\x0a}\x0d\x0a}\x0d\x0a//打印出每一步的路径\x0d\x0avoid move(char a, char c)\x0d\x0a{\x0d\x0aprintf(" %c-->%c\n", a, c)\x0d\x0a}\x0d\x0a\x0d\x0aint main(void)\x0d\x0a{\x0d\x0aint n, time = 0\x0d\x0aprintf("请输入汉诺塔的盘数:")\x0d\x0ascanf("%d", &n)\x0d\x0aprintf("%d个盘的汉诺塔移动方法是：", n)\x0d\x0aprintf("\n")\x0d\x0ahanoi(n, 'A', 'B', 'C', &time)\x0d\x0aprintf("移动了%d次\n", time)\x0d\x0asystem("pause")\x0d\x0areturn 0\x0d\x0a}\x0d\x0a\x0d\x0a非递归算法：\x0d\x0a#include \x0d\x0a\x0d\x0a#define MAXSTACK 10 /* 栈的最大深度 */\x0d\x0a\x0d\x0aint c = 1/* 一个全局变量，表示目前移动的步数 */\x0d\x0a\x0d\x0astruct hanoi { /* 存储汉诺塔的结构，包括盘的数目和三个盘的名称 */\x0d\x0aint n\x0d\x0achar x, y, z\x0d\x0a}\x0d\x0a\x0d\x0avoid move(char x, int n, char y) /* 移动函数，表示把某个盘从某根针移动到另一根针 */\x0d\x0a{\x0d\x0aprintf("%d->%d from %c ->%c\n", c++, n, x, y)\x0d\x0a}\x0d\x0a\x0d\x0avoid hanoi(int n, char x, char y, char z) /* 汉诺塔的递归算法 */\x0d\x0a{\x0d\x0aif (1 == n)\x0d\x0amove(x, 1, z)\x0d\x0aelse {\x0d\x0ahanoi(n - 1, x, z, y)\x0d\x0amove(x, n, z)\x0d\x0ahanoi(n - 1, y, x, z)\x0d\x0a}\x0d\x0a}\x0d\x0a\x0d\x0avoid push(struct hanoi *p, int top, char x, char y, char z,int n)\x0d\x0a{\x0d\x0ap[top+1].n = n - 1\x0d\x0ap[top+1].x = x\x0d\x0ap[top+1].y = y\x0d\x0ap[top+1].z = z\x0d\x0a}\x0d\x0a\x0d\x0avoid unreverse_hanoi(struct hanoi *p) /* 汉诺塔的非递归算法 */\x0d\x0a{\x0d\x0aint top = 0\x0d\x0a\x0d\x0awhile (top >= 0) {\x0d\x0awhile (p[top].n >1) { /* 向左走到尽头 */\x0d\x0a push(p, top, p[top].x, p[top].z, p[top].y, p[top].n)\x0d\x0a top++\x0d\x0a}\x0d\x0aif (p[top].n == 1) { /* 叶子结点 */\x0d\x0a move(p[top].x, 1, p[top].z)\x0d\x0a top--\x0d\x0a}\x0d\x0aif (top >= 0) { /* 向右走一步 */\x0d\x0a move(p[top].x, p[top].n, p[top].z)\x0d\x0a top--\x0d\x0a push(p, top, p[top+1].y, p[top+1].x, p[top+1].z, p[top+1].n)\x0d\x0a top++\x0d\x0a}\x0d\x0a}\x0d\x0a}\x0d\x0a\x0d\x0aint main(void)\x0d\x0a{\x0d\x0aint i\x0d\x0aprintf("递归:\n")\x0d\x0ahanoi(3, 'x', 'y', 'z')\x0d\x0aprintf("非递归:\n")\x0d\x0astruct hanoi p[MAXSTACK]\x0d\x0ac = 1\x0d\x0ap[0].n = 3\x0d\x0ap[0].x = 'x', p[0].y = 'y', p[0].z = 'z'\x0d\x0aunreverse_hanoi(p)\x0d\x0a\x0d\x0areturn 0\x0d\x0a}

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