c语言92

㈠ c璇瑷 杞涔夊瓧绗︿腑 鍑虹幇\8锛\97 锛\992浠ｈ〃浠涔 鍗犲嚑涓瀛楄妭 鎴戠煡阆\ddd鍜\xdd

8锛岃〃绀鍏杩涘埗镄8锛堟崲绠椾负鍗佽繘鍒灏辨槸8锛鍗佸叚杩涘埗镄8锛夌殑ASCII镰佹墍瀵瑰簲镄勭﹀彿锛屽嵆阃镙硷纴涔熷氨鏄灏嗗綋鍓崭綅缃绉诲埌鍓崭竴鍒楋绂

镊充簬97鍜孪992锛屾垜璁や负鏄链夐梾棰樼殑锛屽洜涓鸿浆瀛愬瓧绗﹀彧链夊叓杩涘埗鍜屽崄鍏杩涘埗镄勮〃绀猴纴鍏杩涘埗鏄鐩存帴甯︽暟瀛楃殑锛屽崄鍏杩涘埗鐢▁琛ㄧず锛屾墍浠ddd涓璬dd琛ㄧず镄勬暟瀛楀簲璇ユ槸鍏杩涘埗锛屼篃灏变笉鍙鑳藉嚭鐜板ぇ浜7镄勬暟瀛楋纴xdd琛ㄧず镄勬槸鍗佸叚杩涘埗锛屽悇浣岖殑鏁板瓧鏄浠0~F銆

褰撶劧锛屽傛灉鍙鐪嫔瓧鑺傜殑璇濓纴闾8锛孪97 锛孪992閮藉簲璇ユ槸鍙鍗犱竴涓瀛楄妭镄勚

鍏蜂綋琛ㄧず浠涔堬纴寤鸿鍐欑▼搴忔妸97 锛孪992璧嫔肩粰涓涓猚har鍨嫔彉閲忥纴铹跺悗鍐嶆妸浠栨墦鍑烘潵锛屽悓镞朵篃鎶娄粬镄勬暟鍊兼墦鍗板嚭𨱒ワ纴鐪嬭兘钖︽墦鍗帮纴涓鑸搴旇ユ槸浼氭湁锻婅︾殑鎴栬呭帇镙圭紪璇戜笉阃氲繃銆

鎴戣繖杈圭粰浣犺瘯浜嗕笅锛

浠ｇ爜濡备笅锛

#include <stdio.h>

int main()

{

char temp1, temp2;

temp1 = 97;

temp2 = 992;

printf("temp1=%c,value=%d ",temp1,temp1);

printf("temp2=%c,value=%d ",temp2,temp2);

return 0;

}

缂栬疟涓嶉氲繃锛屽备笅锲撅细

镓浠ワ纴浣犺繖涓镶瀹氭槸鍐欓敊浜嗐

㈡ char s='\92'鐢%c杈揿嚭涓轰粈涔堣緭鍑9锛

鍦╟璇瑷涓锛屸渃har钬濊〃绀哄瓧绗﹀瀷鏁版嵁锛屸榎92钬欐槸杞涔夊瓧绗︼纴琛ㄧず鍏杩涘埗鏁板瓧钬92钬椤嵆鍗佽繘鍒舵暟57瀵瑰簲镄𪞝SC||镰佺殑鍊硷纴镙规嵁ASC||琛ㄥ彲鐭ラ亾锛屽崄杩涘埗鏁57瀵瑰簲镄𪞝SC||镰佷负钬9钬欍傜敤钬%c"杈揿嚭锛屽嵆鐢ㄥ瓧绗﹀瀷鏁版嵁杈揿嚭锛屾墍浠ヨ緭鍑衡9钬溿

ASC||琛

㈢ 求教c语言回溯法写出八皇后问题的92种解

(1)全排列

将自然数1~n进行排列，共形成n!中排列方式，叫做全排列。

例如3的全排列是：1/2/3、1/3/2、2/1/3、2/3/1、3/1/2、3/2/1，共3!=6种。

(2)8皇后（或者n皇后）

保证8个皇后不能互相攻击，即保证每一横行、每一竖行、每一斜行最多一个皇后。

我们撇开第三个条件，如果每一横行、每一竖行都只有一个皇后。

将8*8棋盘标上坐标。我们讨论其中的一种解法：

- - - - - - - Q

- - - Q - - - -

Q - - - - - - -

- - Q - - - - -

- - - - - Q - -

- Q - - - - - -

- - - - - - Q -

- - - - Q - - -

如果用坐标表示就是：(1,8) (2,4) (3,1) (4,3) (5,6) (6,2) (7,7) (8,5)

将横坐标按次序排列，纵坐标就是8/4/1/3/6/2/7/5。这就是1~8的一个全排列。

我们将1~8的全排列存入输入a[]中(a[0]~a[7])，然后8个皇后的坐标就是(i+1,a[i])，其中i为0~7。

这样就能保证任意两个不会同一行、同一列了。

置于斜行，你知道的，两个点之间连线的斜率绝对值为1或者-1即为同一斜行，充要条件是|x1-x2|=|y1-y2|（两个点的坐标为（x1,y1）(x2,y2)）。我们在输出的时候进行判断，任意两个点如果满足上述等式，则判为失败，不输出。

下面附上代码：添加必要的注释，其中全排列的实现看看注释应该可以看懂：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
intprinted;
//该函数用于画图，这里为了节约空间则略去
//读者只需要将draw(a,k);去掉注释即可画图
voiddraw(int*a,intk)
{
	inti,j;
	for(i=0;i<k;i++)
	{
		printf("	");
		for(j=0;j<k;j++)
			//有皇后输出Q，否则输出-
			if(a[i]-1==j)printf("Q");elseprintf("-");
		printf("
");
	}
	printf("
");

}
//递归实现全排列,a是数组，iStep是位置的测试点，k是皇后的个数，一般等于8
voidSettle(int*a,intiStep,intk)
{	
	inti,j,l,flag=1;
	//如果iStep的数字等于a之前的数字，则存在重复，返回
	for(i=0;i<iStep-1;i++)
		if(a[iStep-1]==a[i])return;
	//如果iStep==k，即递归结束到最后一位，可以验证是否斜行满足
	if(iStep==k)
	{
		//双重循环判断是否斜行满足
		for(j=0;j<k;j++)
			for(l=0;l<k&&l!=j;l++)
				//如果不满足，则flag=0
				if(fabs(j-l)==fabs(a[j]-a[l]))flag=0;
		//如果flag==1，则通过了斜行的所有测试，输出。
		if(flag)
		{
			for(i=0;i<k;i++)
				printf("(%d,%d)",i+1,a[i]);
			printf("
");
			//如果去掉这里的注释可以获得画图，由于空间不够，这里略去
		//	draw(a,k);
			//printed变量计算有多少满足题意的结果，是全局变量
			printed++;
		}
		flag=1;
	}
	//如果未测试至最后末尾，则测试下一位（递归）
	for(i=1;i<=k;i++)
	{
		a[iStep]=i;
		Settle(a,iStep+1,k);
	}
}
voidmain()
{
	int*a;
	intk;
	//输入维数，建立数组
	printf("Enterthesizeofthesquare:");
	scanf("%d",&k);
	a=(int*)calloc(k,sizeof(int));
	//清屏，从iStep=0处进入递归
	system("cls");
	Settle(a,0,k);
	//判断最后是否有结果
	if(!printed)printf("Noanswersaccepted!
");
	elseprintf("%dstatesavailable!
",printed);
}
附输出结果（输入k=8）：
(1,1)(2,5)(3,8)(4,6)(5,3)(6,7)(7,2)(8,4)
(1,1)(2,6)(3,8)(4,3)(5,7)(6,4)(7,2)(8,5)
(1,1)(2,7)(3,4)(4,6)(5,8)(6,2)(7,5)(8,3)
(1,1)(2,7)(3,5)(4,8)(5,2)(6,4)(7,6)(8,3)
(1,2)(2,4)(3,6)(4,8)(5,3)(6,1)(7,7)(8,5)
(1,2)(2,5)(3,7)(4,1)(5,3)(6,8)(7,6)(8,4)
(1,2)(2,5)(3,7)(4,4)(5,1)(6,8)(7,6)(8,3)
(1,2)(2,6)(3,1)(4,7)(5,4)(6,8)(7,3)(8,5)
(1,2)(2,6)(3,8)(4,3)(5,1)(6,4)(7,7)(8,5)
(1,2)(2,7)(3,3)(4,6)(5,8)(6,5)(7,1)(8,4)
(1,2)(2,7)(3,5)(4,8)(5,1)(6,4)(7,6)(8,3)
(1,2)(2,8)(3,6)(4,1)(5,3)(6,5)(7,7)(8,4)
(1,3)(2,1)(3,7)(4,5)(5,8)(6,2)(7,4)(8,6)
(1,3)(2,5)(3,2)(4,8)(5,1)(6,7)(7,4)(8,6)
(1,3)(2,5)(3,2)(4,8)(5,6)(6,4)(7,7)(8,1)
(1,3)(2,5)(3,7)(4,1)(5,4)(6,2)(7,8)(8,6)
(1,3)(2,5)(3,8)(4,4)(5,1)(6,7)(7,2)(8,6)
(1,3)(2,6)(3,2)(4,5)(5,8)(6,1)(7,7)(8,4)
(1,3)(2,6)(3,2)(4,7)(5,1)(6,4)(7,8)(8,5)
(1,3)(2,6)(3,2)(4,7)(5,5)(6,1)(7,8)(8,4)
(1,3)(2,6)(3,4)(4,1)(5,8)(6,5)(7,7)(8,2)
(1,3)(2,6)(3,4)(4,2)(5,8)(6,5)(7,7)(8,1)
(1,3)(2,6)(3,8)(4,1)(5,4)(6,7)(7,5)(8,2)
(1,3)(2,6)(3,8)(4,1)(5,5)(6,7)(7,2)(8,4)
(1,3)(2,6)(3,8)(4,2)(5,4)(6,1)(7,7)(8,5)
(1,3)(2,7)(3,2)(4,8)(5,5)(6,1)(7,4)(8,6)
(1,3)(2,7)(3,2)(4,8)(5,6)(6,4)(7,1)(8,5)
(1,3)(2,8)(3,4)(4,7)(5,1)(6,6)(7,2)(8,5)
(1,4)(2,1)(3,5)(4,8)(5,2)(6,7)(7,3)(8,6)
(1,4)(2,1)(3,5)(4,8)(5,6)(6,3)(7,7)(8,2)
(1,4)(2,2)(3,5)(4,8)(5,6)(6,1)(7,3)(8,7)
(1,4)(2,2)(3,7)(4,3)(5,6)(6,8)(7,1)(8,5)
(1,4)(2,2)(3,7)(4,3)(5,6)(6,8)(7,5)(8,1)
(1,4)(2,2)(3,7)(4,5)(5,1)(6,8)(7,6)(8,3)
(1,4)(2,2)(3,8)(4,5)(5,7)(6,1)(7,3)(8,6)
(1,4)(2,2)(3,8)(4,6)(5,1)(6,3)(7,5)(8,7)
(1,4)(2,6)(3,1)(4,5)(5,2)(6,8)(7,3)(8,7)
(1,4)(2,6)(3,8)(4,2)(5,7)(6,1)(7,3)(8,5)
(1,4)(2,6)(3,8)(4,3)(5,1)(6,7)(7,5)(8,2)
(1,4)(2,7)(3,1)(4,8)(5,5)(6,2)(7,6)(8,3)
(1,4)(2,7)(3,3)(4,8)(5,2)(6,5)(7,1)(8,6)
(1,4)(2,7)(3,5)(4,2)(5,6)(6,1)(7,3)(8,8)
(1,4)(2,7)(3,5)(4,3)(5,1)(6,6)(7,8)(8,2)
(1,4)(2,8)(3,1)(4,3)(5,6)(6,2)(7,7)(8,5)
(1,4)(2,8)(3,1)(4,5)(5,7)(6,2)(7,6)(8,3)
(1,4)(2,8)(3,5)(4,3)(5,1)(6,7)(7,2)(8,6)
(1,5)(2,1)(3,4)(4,6)(5,8)(6,2)(7,7)(8,3)
(1,5)(2,1)(3,8)(4,4)(5,2)(6,7)(7,3)(8,6)
(1,5)(2,1)(3,8)(4,6)(5,3)(6,7)(7,2)(8,4)
(1,5)(2,2)(3,4)(4,6)(5,8)(6,3)(7,1)(8,7)
(1,5)(2,2)(3,4)(4,7)(5,3)(6,8)(7,6)(8,1)
(1,5)(2,2)(3,6)(4,1)(5,7)(6,4)(7,8)(8,3)
(1,5)(2,2)(3,8)(4,1)(5,4)(6,7)(7,3)(8,6)
(1,5)(2,3)(3,1)(4,6)(5,8)(6,2)(7,4)(8,7)
(1,5)(2,3)(3,1)(4,7)(5,2)(6,8)(7,6)(8,4)
(1,5)(2,3)(3,8)(4,4)(5,7)(6,1)(7,6)(8,2)
(1,5)(2,7)(3,1)(4,3)(5,8)(6,6)(7,4)(8,2)
(1,5)(2,7)(3,1)(4,4)(5,2)(6,8)(7,6)(8,3)
(1,5)(2,7)(3,2)(4,4)(5,8)(6,1)(7,3)(8,6)
(1,5)(2,7)(3,2)(4,6)(5,3)(6,1)(7,4)(8,8)
(1,5)(2,7)(3,2)(4,6)(5,3)(6,1)(7,8)(8,4)
(1,5)(2,7)(3,4)(4,1)(5,3)(6,8)(7,6)(8,2)
(1,5)(2,8)(3,4)(4,1)(5,3)(6,6)(7,2)(8,7)
(1,5)(2,8)(3,4)(4,1)(5,7)(6,2)(7,6)(8,3)
(1,6)(2,1)(3,5)(4,2)(5,8)(6,3)(7,7)(8,4)
(1,6)(2,2)(3,7)(4,1)(5,3)(6,5)(7,8)(8,4)
(1,6)(2,2)(3,7)(4,1)(5,4)(6,8)(7,5)(8,3)
(1,6)(2,3)(3,1)(4,7)(5,5)(6,8)(7,2)(8,4)
(1,6)(2,3)(3,1)(4,8)(5,4)(6,2)(7,7)(8,5)
(1,6)(2,3)(3,1)(4,8)(5,5)(6,2)(7,4)(8,7)
(1,6)(2,3)(3,5)(4,7)(5,1)(6,4)(7,2)(8,8)
(1,6)(2,3)(3,5)(4,8)(5,1)(6,4)(7,2)(8,7)
(1,6)(2,3)(3,7)(4,2)(5,4)(6,8)(7,1)(8,5)
(1,6)(2,3)(3,7)(4,2)(5,8)(6,5)(7,1)(8,4)
(1,6)(2,3)(3,7)(4,4)(5,1)(6,8)(7,2)(8,5)
(1,6)(2,4)(3,1)(4,5)(5,8)(6,2)(7,7)(8,3)
(1,6)(2,4)(3,2)(4,8)(5,5)(6,7)(7,1)(8,3)
(1,6)(2,4)(3,7)(4,1)(5,3)(6,5)(7,2)(8,8)
(1,6)(2,4)(3,7)(4,1)(5,8)(6,2)(7,5)(8,3)
(1,6)(2,8)(3,2)(4,4)(5,1)(6,7)(7,5)(8,3)
(1,7)(2,1)(3,3)(4,8)(5,6)(6,4)(7,2)(8,5)
(1,7)(2,2)(3,4)(4,1)(5,8)(6,5)(7,3)(8,6)
(1,7)(2,2)(3,6)(4,3)(5,1)(6,4)(7,8)(8,5)
(1,7)(2,3)(3,1)(4,6)(5,8)(6,5)(7,2)(8,4)
(1,7)(2,3)(3,8)(4,2)(5,5)(6,1)(7,6)(8,4)
(1,7)(2,4)(3,2)(4,5)(5,8)(6,1)(7,3)(8,6)
(1,7)(2,4)(3,2)(4,8)(5,6)(6,1)(7,3)(8,5)
(1,7)(2,5)(3,3)(4,1)(5,6)(6,8)(7,2)(8,4)
(1,8)(2,2)(3,4)(4,1)(5,7)(6,5)(7,3)(8,6)
(1,8)(2,2)(3,5)(4,3)(5,1)(6,7)(7,4)(8,6)
(1,8)(2,3)(3,1)(4,6)(5,2)(6,5)(7,7)(8,4)
(1,8)(2,4)(3,1)(4,3)(5,6)(6,2)(7,7)(8,5)
92statesavailable!

㈣ C语言输出99乘法表

1、首先使用vs2017新建一个c语言的文件，引入头文件并写好main主函数：

㈤ 单片机数码管显示0到999c语言程序怎么编

#include<reg51.h>

unsigned char xs_d[]={0xc0,0xf9,0xa4,0xb0,0x99,0x92,0x82,0xf8,0x80,0x90};

unsigned int time=0,s,sz;delay(unsigned int k)
{

unsigned int i,j;

for(i=0;i<k;i++)

for(j=0;j<125;j++);

}INT_0()interrupt 0

{ delay(10);

if(INT0==0){sz++;<br> if(sz>2){sz=0;}

}
}

void T0_int()interrupt 1
{

TH0=(65535-50000)/256;//设置初值

TL0=(65535-50000)%256;

s++;
if(s>20){s=0;<br> if(sz==1)time++;<br> if(time>999){time=0;}

if(sz==0){time=0;//清零<br> }

(5)c语言92扩展阅读：

运算器由运算部件——算术逻辑单元（Arithmetic & Logical Unit，简称ALU）、累加器和寄存器等几部分组成。

ALU的作用是把传来的数据进行算术或逻辑运算，输入来源为两个8位数据，分别来自累加器和数据寄存器。ALU能完成对这两个数据进行加、减、与、或、比较大小等操作，最后将结果存入累加器。

运算器有两个功能：

(1) 执行各种算术运算。

(2) 执行各种逻辑运算，并进行逻辑测试，如零值测试或两个值的比较。

运算器所执行全部操作都是由控制器发出的控制信号来指挥的，并且，一个算术操作产生一个运算结果，一个逻辑操作产生一个判决。