void plus(char *a, char *b, char *c)
{
	int al,bl,len;
	int d[1000],e[1000],f[1000];//用了三個陣列去做，覺得很麻煩，又沒想到簡便的方法。
	memset(d,0,sizeof(d));
	memset(e,0,sizeof(e));
	memset(f,0,sizeof(f));
	char tmpa[1000] = {NULL},tmpb[1000] = {NULL};
	int tmp = 0;
	int ff = 1;//前導零的消去
	for (int i = 0; a[i]; i++)	
	{							
		if(ff && a[i] != '0')ff = 0;
		if(!ff)tmpa[tmp++] = a[i];
	}tmpa[tmp] = '\0';			
	ff = 1,tmp = 0;				
	for (int i = 0; b[i]; i++)	
	{							
		if(ff && b[i]!='0')ff = 0;
		if(!ff)tmpb[tmp++] = b[i];
	}tmpb[tmp] = '\0';			
	al = strlen(tmpa);			
	bl = strlen(tmpb);	
	len = al>bl?al:bl;
	for (int i = 0; i < al; i++)	
		d[i] = tmpa[al-i-1]-'0';//反轉儲存，把字元型變成int型
	for (int i = 0; i < bl; i++)
		e[i] = tmpb[bl-i-1]-'0';
	for (int i = 0; i < len; i++)	{//模擬加法
		f[i] += d[i] + e[i];
		if(f[i]>=10)		{//進位
			f[i+1]++;
			f[i] -= 10;
		}
	}
	if(f[len])len++;//可能多一位
	for (int i = 0; i < len; i++)	
		c[i] = f[len-i-1]+'0';
	c[len] = '\0';
}
void minus(char *a, char *b, char *c)
{
	int al = 0,bl = 0;
	char d[1000]={NULL},e[1000] = {NULL};
	int flag = 0;//判斷正負，flag = 0表示結果為正，同理1。
	char tmpa[1000] = {NULL},tmpb[1000] = {NULL};
	int tmp = 0;
	int f = 1;//前導零的消去
	for (int i = 0; a[i]; i++)				//*******************************************************
	{										//*比較減法的結果是正還是負，不能直接用字串的長度		*
		if(f && a[i] != '0')f = 0;			//*因為有前導零，可以使得數值較小的數 佔用的 長度更長。 *
		if(!f)tmpa[tmp++] = a[i];			//*														*
	}tmpa[tmp] = '\0';						//*														*
	f = 1,tmp = 0;							//*														*
	for (int i = 0; b[i]; i++)				//*														*
	{										//*														*
		if(f && b[i]!='0')f = 0;			//*														*
		if(!f)tmpb[tmp++] = b[i];			//*														*
	}tmpb[tmp] = '\0';						//*														*
	al = strlen(tmpa);						//*														*
	bl = strlen(tmpb);						//*******************************************************/	
	if(strcmp(tmpa,tmpb) == 0){//首先判斷相等的情況
		c[0] = '0';
		c[1] = '\0';
		return ;
	}
	if(al<bl || (al == bl&&strcmp(tmpa,tmpb) < 0))		flag = 1;//為負
	if(!flag){//為正 時。
		for (int i = 0; i <= al; i++)//把位數少的補上前導零，如1010 - 98變成1010 - 0098.
		{
			if(i<al-bl)	d[i] = '0';
			else d[i] = tmpb[i-al+bl];
		}
		for (int i = al-1; i >= 0; i--)//模擬減法
		{
			if(tmpa[i] >= d[i])
				tmpa[i] -= d[i]-'0';
			else{
				tmpa[i-1] -= 1;//退位
				tmpa[i] += 10-d[i]+'0';
			}
		}
		int tmp = 1,x = 0;//用於處理結果中的前導零
		for (int i = 0; i < al; i++)
		{
			if(tmpa[i] == '0' && tmp)
				continue;
			else{
				c[x++] = tmpa[i];
				tmp = 0;
			}
		}c[x] = '\0';//結束
	}else{//為負 時
		//好像能直接呼叫一下minus(b,a,c)，再加個減號就行了？？？
		for (int i = 0; i <= bl; i++)//同上面一樣~~
		{
			if(i<bl-al)	e[i] = '0';
			else e[i] = tmpa[i-bl+al]; 
		}
		for (int i = bl-1; i >= 0; i--)
		{
			if(tmpb[i] >= e[i])
				tmpb[i] -= e[i]-'0';
			else{
				tmpb[i-1] -= 1;
				tmpb[i] += 10-e[i]+'0';
			}
		}
		int tmp = 1,x = 1;		
		c[0] = '-';//加個減號就行
		for (int i = 0; i < bl; i++)
		{
			if(tmpb[i] == '0' && tmp)
				continue;
			else{
				c[x++] = tmpb[i];
				tmp = 0;
			}
		}c[x] = '\0';		
	}
}