c語言92

㈠ c璇璦 杞涔夊瓧絎︿腑 鍑虹幇\8錛\97 錛\992浠ｈ〃浠涔 鍗犲嚑涓瀛楄妭 鎴戠煡閬\ddd鍜\xdd

8錛岃〃紺鍏榪涘埗鐨8錛堟崲綆椾負鍗佽繘鍒灝辨槸8錛鍗佸叚榪涘埗鐨8錛夌殑ASCII鐮佹墍瀵瑰簲鐨勭﹀彿錛屽嵆閫鏍礆紝涔熷氨鏄灝嗗綋鍓嶄綅緗縐誨埌鍓嶄竴鍒楋紱

鑷充簬97鍜孿992錛屾垜璁や負鏄鏈夐棶棰樼殑錛屽洜涓鴻漿瀛愬瓧絎﹀彧鏈夊叓榪涘埗鍜屽嶮鍏榪涘埗鐨勮〃紺猴紝鍏榪涘埗鏄鐩存帴甯︽暟瀛楃殑錛屽嶮鍏榪涘埗鐢▁琛ㄧず錛屾墍浠ddd涓璬dd琛ㄧず鐨勬暟瀛楀簲璇ユ槸鍏榪涘埗錛屼篃灝變笉鍙鑳藉嚭鐜板ぇ浜7鐨勬暟瀛楋紝xdd琛ㄧず鐨勬槸鍗佸叚榪涘埗錛屽悇浣嶇殑鏁板瓧鏄浠0~F銆

褰撶劧錛屽傛灉鍙鐪嬪瓧鑺傜殑璇濓紝閭8錛孿97 錛孿992閮藉簲璇ユ槸鍙鍗犱竴涓瀛楄妭鐨勩

鍏蜂綋琛ㄧず浠涔堬紝寤鴻鍐欑▼搴忔妸97 錛孿992璧嬪肩粰涓涓猚har鍨嬪彉閲忥紝鐒跺悗鍐嶆妸浠栨墦鍑烘潵錛屽悓鏃朵篃鎶婁粬鐨勬暟鍊兼墦鍗板嚭鏉ワ紝鐪嬭兘鍚︽墦鍗幫紝涓鑸搴旇ユ槸浼氭湁鍛婅︾殑鎴栬呭帇鏍圭紪璇戜笉閫氳繃銆

鎴戣繖杈圭粰浣犺瘯浜嗕笅錛

浠ｇ爜濡備笅錛

#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;

}

緙栬瘧涓嶉氳繃錛屽備笅鍥撅細

鎵浠ワ紝浣犺繖涓鑲瀹氭槸鍐欓敊浜嗐

㈡ char s='\92'鐢%c杈撳嚭涓轟粈涔堣緭鍑9錛

鍦╟璇璦涓錛屸渃har鈥濊〃紺哄瓧絎﹀瀷鏁版嵁錛屸榎92鈥欐槸杞涔夊瓧絎︼紝琛ㄧず鍏榪涘埗鏁板瓧鈥92鈥欏嵆鍗佽繘鍒舵暟57瀵瑰簲鐨凙SC||鐮佺殑鍊礆紝鏍規嵁ASC||琛ㄥ彲鐭ラ亾錛屽嶮榪涘埗鏁57瀵瑰簲鐨凙SC||鐮佷負鈥9鈥欍傜敤鈥%c"杈撳嚭錛屽嵆鐢ㄥ瓧絎﹀瀷鏁版嵁杈撳嚭錛屾墍浠ヨ緭鍑衡9鈥溿

ASC||琛

㈢ 求教c語言回溯法寫出八皇後問題的92種解

(1)全排列

將自然數1~n進行排列，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種。

(2)8皇後（或者n皇後）

保證8個皇後不能互相攻擊，即保證每一橫行、每一豎行、每一斜行最多一個皇後。

我們撇開第三個條件，如果每一橫行、每一豎行都只有一個皇後。

將8*8棋盤標上坐標。我們討論其中的一種解法：

- - - - - - - Q

- - - Q - - - -

Q - - - - - - -

- - Q - - - - -

- - - - - Q - -

- Q - - - - - -

- - - - - - Q -

- - - - Q - - -

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

將橫坐標按次序排列，縱坐標就是8/4/1/3/6/2/7/5。這就是1~8的一個全排列。

我們將1~8的全排列存入輸入a[]中(a[0]~a[7])，然後8個皇後的坐標就是(i+1,a[i])，其中i為0~7。

這樣就能保證任意兩個不會同一行、同一列了。

置於斜行，你知道的，兩個點之間連線的斜率絕對值為1或者-1即為同一斜行，充要條件是|x1-x2|=|y1-y2|（兩個點的坐標為（x1,y1）(x2,y2)）。我們在輸出的時候進行判斷，任意兩個點如果滿足上述等式，則判為失敗，不輸出。

下面附上代碼：添加必要的注釋，其中全排列的實現看看注釋應該可以看懂：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
intprinted;
//該函數用於畫圖，這里為了節約空間則略去
//讀者只需要將draw(a,k);去掉注釋即可畫圖
voiddraw(int*a,intk)
{
	inti,j;
	for(i=0;i<k;i++)
	{
		printf("	");
		for(j=0;j<k;j++)
			//有皇後輸出Q，否則輸出-
			if(a[i]-1==j)printf("Q");elseprintf("-");
		printf("
");
	}
	printf("
");

}
//遞歸實現全排列,a是數組，iStep是位置的測試點，k是皇後的個數，一般等於8
voidSettle(int*a,intiStep,intk)
{	
	inti,j,l,flag=1;
	//如果iStep的數字等於a之前的數字，則存在重復，返回
	for(i=0;i<iStep-1;i++)
		if(a[iStep-1]==a[i])return;
	//如果iStep==k，即遞歸結束到最後一位，可以驗證是否斜行滿足
	if(iStep==k)
	{
		//雙重循環判斷是否斜行滿足
		for(j=0;j<k;j++)
			for(l=0;l<k&&l!=j;l++)
				//如果不滿足，則flag=0
				if(fabs(j-l)==fabs(a[j]-a[l]))flag=0;
		//如果flag==1，則通過了斜行的所有測試，輸出。
		if(flag)
		{
			for(i=0;i<k;i++)
				printf("(%d,%d)",i+1,a[i]);
			printf("
");
			//如果去掉這里的注釋可以獲得畫圖，由於空間不夠，這里略去
		//	draw(a,k);
			//printed變數計算有多少滿足題意的結果，是全局變數
			printed++;
		}
		flag=1;
	}
	//如果未測試至最後末尾，則測試下一位（遞歸）
	for(i=1;i<=k;i++)
	{
		a[iStep]=i;
		Settle(a,iStep+1,k);
	}
}
voidmain()
{
	int*a;
	intk;
	//輸入維數，建立數組
	printf("Enterthesizeofthesquare:");
	scanf("%d",&k);
	a=(int*)calloc(k,sizeof(int));
	//清屏，從iStep=0處進入遞歸
	system("cls");
	Settle(a,0,k);
	//判斷最後是否有結果
	if(!printed)printf("Noanswersaccepted!
");
	elseprintf("%dstatesavailable!
",printed);
}
附輸出結果（輸入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語言輸出99乘法表

1、首先使用vs2017新建一個c語言的文件，引入頭文件並寫好main主函數：

㈤ 單片機數碼管顯示0到999c語言程序怎麼編

#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;//設置初值

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語言92擴展閱讀：

運算器由運算部件——算術邏輯單元（Arithmetic & Logical Unit，簡稱ALU）、累加器和寄存器等幾部分組成。

ALU的作用是把傳來的數據進行算術或邏輯運算，輸入來源為兩個8位數據，分別來自累加器和數據寄存器。ALU能完成對這兩個數據進行加、減、與、或、比較大小等操作，最後將結果存入累加器。

運算器有兩個功能：

(1) 執行各種算術運算。

(2) 執行各種邏輯運算，並進行邏輯測試，如零值測試或兩個值的比較。

運算器所執行全部操作都是由控制器發出的控制信號來指揮的，並且，一個算術操作產生一個運算結果，一個邏輯操作產生一個判決。