📄 24allfinished_none2.dat
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#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<time.h>
#define MAX 13
#define OK 1
#define TRUE 1
#define FALSE 0
#define ERROR 0
#define OVERFLOW -2
#define OPSETSIZE 7
#define STACK_INIF_SIZE 50
#define STACKINCREMENT 10
typedef int status;
int number[2][4];
enum
{
eNumber = 0, //操作数
eOperator = 1 //算子
};
int oper[7]={43,45,42,47,40,41,35};
typedef struct sqlist{
int bol;//bol is 0 when num_ch is a number;bol is 1 when the num_ch is a oprater
int num_ch;
struct sqlist *next;
}sqlist;
typedef struct sqstack{
int *base;
int *top;
int stacksize;
}sqstack;
unsigned char Prior[7][7] = { // 表3.1 算符间的优先关系
'>','>','<','<','<','>','>',
'>','>','<','<','<','>','>',
'>','>','>','>','<','>','>',
'>','>','>','>','<','>','>',
'<','<','<','<','<','=',' ',
'>','>','>','>',' ','>','>',
'<','<','<','<','<',' ','='
};
char OPSET[OPSETSIZE]={'+' , '-' , '*' , '/' ,'(' , ')' , '#'};
status init_sq(sqlist *l){//初始化链表
l=(sqlist*)malloc(sizeof(sqlist));
if(l==NULL){
exit(OVERFLOW);
}
l->next=NULL;
return OK;
}
status insert_sq(sqlist **p,int e,int bl){//链表插入操作
sqlist *q;
q=(sqlist*)malloc(sizeof(sqlist));
q->num_ch=e;
q->bol=bl;
q->next=NULL;
(*p)->next=q;
(*p)=(*p)->next;
// printf("insert %d ,%dis succssed!\n",e,bl);
return OK;
}
int check(sqlist l)//保证输入的数字是给出的四个数字
{
int right=1,find=0,i;
sqlist *q=&l;
q=q->next ;
for (;q->next!=NULL;q=q->next){
if(q->bol==1){
if(q->num_ch <=39||q->num_ch>57||q->num_ch==44||q->num_ch==46){
right=0;
printf("%c不是有效的运算符!\n");
}
}
else {
find=0;
for(i=0;i<4;i++){
if(number[1][i]==0&&number[0][i]==q->num_ch ){
number[1][i]=1;
find=1;
break;
}
}
if(find==0){
printf("%d 不在给出的四个数字中!\n",q->num_ch );
right=0;
}
}
}//end for
for (i=0;i<4;i++){
if(number[1][i]==0){
printf("%d没有用上!\n",number[0][i]);
right=0;
}
}
return right;
}
int chang(char *s,sqlist *l){//将用户的输入转化为单链表
int t=0;
unsigned int i=0;
int bl,ch;
int a1,a2,a;
sqlist *p=l;
for (;i<strlen(s);i++){
if(s[i]>47&&s[i]<58&&t==0){
a1=(int)s[i]-48;
t++;
}
else if(s[i]>47&&s[i]<58&&t==1){
a2=(int)s[i]-48;
a=a1*10+a2;
t++;
}
else if(s[i]<48&&s[i]>39&&s[i]!=44&&s[i]!=46){
if(t==1){
bl=0;
insert_sq(&p,a1,bl);
t=0;
}
else if(t==2){
bl=0;
insert_sq(&p,a,bl);
t=0;
}
bl=1;
ch=(int)s[i];
insert_sq(&p,ch,bl);
t=0;
}
else {
printf("%c不是有效的运算符!\n",s[i]);
}
} //end for
i=strlen(s)-1;
if(s[i]>47&&s[i]<58){
if(s[i-1]>47&&s[i-1]<58){
bl=0;
insert_sq(&p,a,bl);
}
else {
bl=0;
insert_sq(&p,a1,bl);
}
}
bl=1;
a=35;
insert_sq(&p,a,bl);
// printf("chang is OK\n");
return (check(*l));
}
int Operate(int a,int theta, int b) {//计算
// printf("a=%d,theta=%c,b=%d\n",a,theta,b);
switch(theta) {
case 43: return a+b;
case 45: return a-b;
case 42: return a*b;
case 47:
{
if(b==0){
return -2000;
}
if (a%b==0){
return a/b;
}
else {//printf("不能为小数\n");
return -10000;
}
}
default : return 0;
}
}
int ReturnOpOrd(char op,char* TestOp) {//被char precede(char Aop, char Bop)所调用来求优先级
int i;
for(i=0; i< OPSETSIZE; i++) {
if (op == TestOp[i]) return i;
}
return 0;
}
char precede(char Aop, char Bop) {
return Prior[ReturnOpOrd(Aop,OPSET)][ReturnOpOrd(Bop,OPSET)];
}
status initstack(sqstack *s){
(s)->base = (int*)malloc(STACK_INIF_SIZE*sizeof(int));
if((s)->base==NULL) exit(OVERFLOW);
(s)->top=(s)->base;
(s)->stacksize = STACK_INIF_SIZE;
// printf("栈初始化完成!\n");
return OK;
}
int gettop(sqstack *s){
int e;
if(s->top==s->base){
printf("栈空,无法取得栈顶元素!\n");
return 0;
}
e=*(s->top-1);
// printf("取得栈顶元素: %d \n",e);
return e;
}
status push(sqstack *s,int e){
if(s->top-s->base>=s->stacksize){
s->base=(int*)realloc(s->base,(s->stacksize+STACKINCREMENT)*sizeof(int));
if(!s->base) exit(OVERFLOW);
s->stacksize+= STACKINCREMENT;
}
*(s->top++)=e;
// printf("把 %d 压栈 is OK\n",e);
return OK;
}
status pop(sqstack *s,int *e){
if(s->top==s->base){
printf("栈空,出栈错误!\n");
return ERROR;
}
*e=*(--s->top);
// printf(" %d出栈成功 !\n",*e);
return OK;
}
int EvaluateExpression(char* MyExpression) { // 算法3.4
// 算术表达式求值的算符优先算法。
// 设OPTR和&&OPND分别为运算符栈和运算数栈，OP为运算符集合。
int result;
sqstack OPTR; // 运算符栈，字符元素
sqstack OPND; // 运算数栈，实数元素
int c,bl,a,b,theta,top;
sqlist *q,l;
char *s=MyExpression;
init_sq(&l);
if(chang(s,&l)!=0){
q=&l;
initstack(&OPTR);
push(&OPTR, 35);
initstack (&OPND);
q=q->next;
c=q->num_ch;
bl=q->bol;
while ((c!= 35 || gettop(&OPTR)!=35)){
if (bl!=1) {
push(&OPND, c);
q=q->next;
c=q->num_ch;
bl=q->bol;
} // 不是运算符则进栈
else {
top=gettop(&OPTR);
// printf("top %c",top);
switch (precede(top, c)) {
case '<': // 栈顶元素优先权低
push(&OPTR, c);
q=q->next;
c=q->num_ch;
bl=q->bol;
break;
case '=': // 脱括号并接收下一字符
pop(&OPTR, &c);
q=q->next;
c=q->num_ch;
bl=q->bol;
break;
case '>': // 退栈并将运算结果入栈
pop(&OPTR, &theta);
pop(&OPND, &b);
pop(&OPND, &a);
// printf("q->num_ch is %d\n",q->num_ch);
push(&OPND, Operate(a, theta, b));
break;
default :
printf("没有这个运算符!\n");
return 0;
} // switch
}//else
} // while
result=gettop(&OPND);
return result;
}
else {
printf("你的输入有错误!\n");
return 0;
}
} // EvaluateExpression
int randomm()//产生四个随机数
{
int i=0;
srand((unsigned)time(NULL));
for (;i<4;i++){
number[0][i]=0;
number[0][i]=rand();
number[0][i]%=13;
number[0][i]++;
number[1][i]=0;
}
return number[2][4];
}
int CalcOneExpress(int expression[][2])
{
// 算术表达式求值的算符优先算法。
// 设OPTR和&&OPND分别为运算符栈和运算数栈，OP为运算符集合。
int index=0,result,c,theta,a,b;
sqstack OPTR; // 运算符栈，字符元素
sqstack OPND; // 运算数栈，实数元素
initstack(&OPTR);
push(&OPTR, 35);
initstack (&OPND);
c=expression[index][0];
while (c!= 35 || gettop(&OPTR)!=35){
if (expression[index][1]!=1) {
push(&OPND, c);
index++;
c=expression[index][0];
} // 不是运算符则进栈
else {
switch (precede(gettop(&OPTR), c)) {
case '<':
// 栈顶元素优先权低
push(&OPTR, c);
index++;
c=expression[index][0];
break;
case '=': // 脱括号并接收下一字符
pop(&OPTR, &c);
index++;
c=expression[index][0];
break;
case '>': // 退栈并将运算结果入栈
pop(&OPTR, &theta);
pop(&OPND, &b);
pop(&OPND, &a);
push(&OPND, Operate(a, theta, b));
break;
default :
printf("没有这个运算符\n");
return 0;
} // switch
}//else
} // while
result=gettop(&OPND);
return result;
} // EvaluateExpression
int Equal24(int n)
{
if(n==24){
// printf("the result is %d\n",n);
return TRUE;
}
else return FALSE;
}
//括号的几种情况
//1 无括号
//2 (a b) c d 同a b (c d), 下省略
//3 (a b c) d
//4 a (b c) d
//5 (a b) (c d)
//6 ((a b) c) d
//7 (a (b c)) d
int CalcArray1(int iNumInput[2][4])
{
// a * b * c * d //7 number
int expression[8][2],ii,jj,kk;
int i,j,k,l,dRes;
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
if(j==i)
{
continue;
}
for(k=0;k<4;k++)
{
if(k==i||k==j)
{
continue;
}
for(l=0;l<4;l++)
{
if(l==i||l==j||l==k)
{
continue;
}
expression[0][0]=iNumInput[0][i];
expression[2][0]=iNumInput[0][j];
expression[4][0]=iNumInput[0][k];
expression[6][0]=iNumInput[0][l];
expression[0][1]=eNumber;
expression[2][1]=eNumber;
expression[4][1]=eNumber;
expression[6][1]=eNumber;
for (ii=0;ii<4;ii++)
{
for (jj=0;jj<4;jj++)
{
for (kk=0;kk<4;kk++)
{
expression[1][0] = oper[ii];
expression[1][1] = eOperator;
expression[3][0] = oper[jj];
expression[3][1] = eOperator;
expression[5][0] = oper[kk];
expression[5][1] = eOperator;
expression[7][0] = oper[6];
expression[7][1] = eOperator;
dRes = CalcOneExpress(expression);
if(Equal24(dRes))
{
printf("可以这样运算:%d%c%d%c%d%c%d\n",expression[0][0],oper[ii],expression[2][0],oper[jj],expression[4][0],oper[kk],expression[6][0]);
return TRUE;
}
}
}
}//end of for oper
}
}
}
}
return FALSE;
}
int CalcArray2(int iNumInput[2][4])
{
// (a * b) * c * d //9 number
int expression[10][2];
int ii,jj,i,j,k,l,kk;
int dRes;
// printf("CalcArray2\n");
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
if(j==i)
{
continue;
}
for(k=0;k<4;k++)
{
if(k==i||k==j)
{
continue;
}
for(l=0;l<4;l++)
{
if(l==i||l==j||l==k)
{
continue;
}
expression[1][0]=iNumInput[0][i];
expression[3][0]=iNumInput[0][j];
expression[6][0]=iNumInput[0][k];
expression[8][0]=iNumInput[0][l];
expression[1][1]=eNumber;
expression[3][1]=eNumber;
expression[6][1]=eNumber;
expression[8][1]=eNumber;
for (ii=0;ii<4;ii++)
{
for (jj=0;jj<4;jj++)
{
for (kk=0;kk<4;kk++)
{
expression[2][0] = oper[ii];
expression[2][1] = eOperator;
expression[5][0] = oper[jj];
expression[5][1] = eOperator;
expression[7][0] = oper[kk];
expression[7][1] = eOperator;
expression[9][0] = oper[6];
expression[9][1] = eOperator;
expression[0][0] = oper[4];
expression[0][1] = eOperator;
expression[4][0] = oper[5];
expression[4][1] = eOperator;
dRes = CalcOneExpress(expression);
if(Equal24(dRes))
{
printf("可以这样运算:%c%d%c%d%c%c%d%c%d\n",oper[4],expression[1][0],oper[ii],expression[3][0],oper[5],oper[jj],expression[6][0],oper[kk],expression[8][0]);
return TRUE;
}
}
}
}//end of for oper
}
}
}
}
return FALSE;
}
int CalcArray3(int iNumInput[2][4])
{
// (a * b * c) * d //9 number
int expression[10][2];
int ii,jj,i,j,k,l,kk;
int dRes;
// printf("CalcArray3\n");
for(i=0;i<4;i++)
{
for(j=0;j<4;j++)
{
if(j==i)
{
continue;
}
for(k=0;k<4;k++)
{
if(k==i||k==j)
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