main()//m为4，n为5，其它为1，3，11，15，32
{
int m=1,n=1,i,j,z,x,k,y,t,l,zuida;//m=4张，n=5种,x就为它最大的连续的那一个数,b记录已用的张数
int a[10],b[MAX];
for(i=0;i<MAX;i++)
b[i]=0;
printf("please input the m and n: ");
while(m!=0 && n!=0)
{
scanf("%d %d",&m,&n);
printf("please input the value:");
for(i=0;i<n;i++)
scanf("%d",&a[i]);//存储各个数
paixu(a,n);//对输入的几个数排序
for(i=0;i<n;i++)
b[i]=a[i];//第一轮
zuida=jisuan(m,n);//计算总共有多少种可重复的可能,是n的m次方
t=1;
k=0;
l=n;
for(;i<zuida;)
{
t=i;
for(j=0;j<l;j++)
{
b[i]=b[t-(n+k*5)+k]+a[j];
printf("%d ",b[i]);
i++;
if(i==(t+5))
k++;
if(k==5)
{
k=0;
n=n*n;
}
}
}
printf("\n");
z=1;//我记得前面没有改变z的值嘎，为什么这里不给它赋值就不得行呀
y=1;
x=1;
printf("\n%d ",z);
while(y==1)//m为4，n为5，其它为1，3，11，15，32
{
for(i=0;i<zuida && x==1;i++)
{
if(z==b[i])
{
x=0;
z++;
printf("%d ",z);
}
}
if(x==1)
y=0;
x=1;
}
printf("\n\n1-%d\n",z-1);
}
}
//when popping one char in charstack, two nums will be popped.
void pop_pro(stacknum *list_num, stackchar *list_char)
{
int pop_num1,pop_num2,res;
char pop_char1;
pop_num1=pop_num_stack(list_num);//pop the first elem
pop_num2=pop_num_stack(list_num);//pop the second elem
pop_char1=pop_char_stack(list_char);//pop the top operator
res=jisuan( pop_char1, pop_num1, pop_num2);//calcu the res and push it into the numstack
push_num_stack(list_num, res);
}
int main(int argc, char *argv[])
{
double ps[10][14]; int i, j;
double r[] = { 1.5,2,3,4,5,6,9,12,15,20,30,45,60,90 };// AB/2
double p1[] = { 50,60,60,100,250,40,30,20,50,5 };//一层电阻率
double p2[] = { 40,30,20,50,5,50,60,60,100,250 };//二层电阻率
double h[] = { 2,4,6,8,10,2,4,6,8,10 };//层厚度
FILE *fp;
fp = fopen("xxx.txt", "w");
for (i = 0; i<10; i++)
{
for (j = 0; j<14; j++)
{
ps[i][j] = jisuan(r[j], p1[i], p2[i], h[i]);
//fprintf(fp,"%f\t%f\t",r[j]/2/h[i],ps[i][j]/p1[i]);
//printf("%f\t%f\t",r[j]/2/h[i],ps[i][j]/p1[i]);
}
//fprintf(fp,"\n");
//printf("\n");
}
for (j = 0; j<14; j++)
{
for (i = 0; i<10; i++)
{
printf("%f\t%f\t", r[j] / 2 / h[i], ps[i][j] / p1[i]);
fprintf(fp, "%f\t%f\t", r[j] / 2 / h[i], ps[i][j] / p1[i]);
}
fprintf(fp, "\n");
printf("\n");
}
return 0;
}
struct Node* regular_expression_syntax_tree(char *s)
{
int i,j,f=0;
inpp=outpp=0;
for(i=0;;i++)
{
if(!isprint(s[i]) && s[i]!='\t' && s[i]!='\n' && s[i]!='\0')return NULL;
switch(s[i])
{
case '*':
case '\0':
case '|':
while(inpp>0 && priority[in[inpp]]>=priority[s[i]])
if(jisuan(in[inpp--])==-1)return NULL;
in[++inpp]=s[i];
if(s[i]=='|')f=0;
break;
case '(':
if(f)
{
while(inpp>0 && priority[in[inpp]]>=priority['&'])
if(jisuan(in[inpp--])==-1)return NULL;
in[++inpp]='&';
}
in[++inpp]=s[i];
f=0;
break;
case ')':
while(inpp>0 && in[inpp]!='(')
if(jisuan(in[inpp--])==-1)return NULL;
if(inpp--==0)return NULL;
f=1;
break;
case '@':
char regular[16];
for(j=0;s[++i]!='@' && s[i]!='\0' && j<15;j++)regular[j]=s[i];
if(s[i]!='@')return NULL;
regular[j]='\0';
for(j=0;j<regular_table_size;j++)if(strcmp(regular,regulars[j].name)==0)break;
if(j==regular_table_size || regulars[j].tree==NULL)return NULL;
if(f)
{
while(inpp>0 && priority[in[inpp]]>=priority['&'])
if(jisuan(in[inpp--])==-1)return NULL;
in[++inpp]='&';
}
out[++outpp]=copyTree(regulars[j].tree);
f=1;
break;
case '\\':
default:
int trans=0;
if(s[i]=='\\')
{
i++;
trans=1;
}
if(f)
{
while(inpp>0 && priority[in[inpp]]>=priority['&'])
if(jisuan(in[inpp--])==-1)return NULL;
in[++inpp]='&';
}
int t;
if(!trans && s[i]=='$')t='$'+256;
else t=s[i];
struct Node *x=newNode(t,NULL,NULL);
if(trans || s[i]!='$')
{
isinput[s[i]]=1;
x->n=++cc;
C[x->n]=s[i];
}
out[++outpp]=x;
f=1;
break;
}
if(s[i]=='\0')break;
}
if(inpp!=1 || outpp!=1)return NULL;
return out[1];
}
