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/*
题目二
表达式求值问题
给定一个算术表达式,对该表达式进行四则运算。
基本要求:
(1)以栈作为存储结构,一个栈存放操作符,一个栈存放操作数;
(2)按字符串形式输入表达式;
(3)最终结果放在操作数栈中并输出。
测试数据要求:
给定的表达式中,必须有包含两位数的数值。例如,给定的表达式为:(10+2)*6-12/3
*/
#include <stdio.h>
#include <stdlib.h>
#define STACK_INIT_SIZE 100
#define STACKINCREMENT 30
typedef struct{
char *base;
int top;
int stacksize;
}SqStack;
typedef struct{
int *base;
int top;
int stacksize;
}NumStack;
void Init_SqStack(SqStack &S){
S.base=(char *)malloc(STACK_INIT_SIZE*sizeof(char));
if(!S.base){
return ;
}
S.top=0;
S.stacksize=STACK_INIT_SIZE;
}
int SqStackEmpty(SqStack S){
if(S.top==0){
return 1;
}else{
return 0;
}
}
void Push_Sq(SqStack &S,char e){
if(S.top>=S.stacksize){
S.base=(char *)realloc(S.base,(S.stacksize+STACKINCREMENT)*sizeof(char));
if(!S.base){
exit(0);
}
S.stacksize+=STACKINCREMENT;
}
S.base[S.top++]=e;
return ;
}
void Pop_Sq(SqStack &S,char &e){
if(S.top==0){
exit(0);
}
e=S.base[--S.top];
return ;
}
char GetTop_Sq(SqStack S,char &e){
if(S.top==0){
return NULL;
}
e=S.base[S.top-1];
return e;
}
void Init_NumStack(NumStack &S){
S.base=(int *)malloc(STACK_INIT_SIZE*sizeof(int));
if(!S.base){
return ;
}
S.top=0;
S.stacksize=STACK_INIT_SIZE;
}
int NumStackEmpty(NumStack S){
if(S.top==0){
return 1;
}else{
return 0;
}
}
void Push_Num(NumStack &S,int e){
if(S.top>=S.stacksize){
S.base=(int *)realloc(S.base,(S.stacksize+STACKINCREMENT)*sizeof(int));
if(!S.base){
exit(0);
}
S.stacksize+=STACKINCREMENT;
}
S.base[S.top++]=e;
return ;
}
void Pop_Num(NumStack &S,int &e){
if(S.top==0){
exit(0);
}
e=S.base[--S.top];
return ;
}
void GetTop_Num(NumStack S,int &e){
if(S.top==0){
return ;
}
e=S.base[S.top-1];
return ;
}
char Priority(char a,char b){
if(a=='*'||a=='/'){
if(b=='+'||b=='-'){
return '>';
}else if(b=='*'||b=='/'){
return '=';
}else{
return '<';
}
}
if(a=='+'||a=='-'){
if(b=='+'||b=='-'){
return '=';
}
else{
return '<';
}
}
}
int calculate(int num1,char x,int num2){
if(x=='+'){
return num1+num2;
}
else if(x=='-'){
return num1-num2;
}
else if(x=='*'){
return num1*num2;
}
else if(x=='/'){
return num1/num2;
}
}
void Init_Storage(char a[]){
NumStack Num;
SqStack Oper;
Init_SqStack(Oper);
Init_NumStack(Num);
Push_Sq(Oper,'#');
int i=0;
char x,jud;
int num1,num2,result=0;
while(a[i]!='\0'){
if(a[i]>='0'&&a[i]<='9'){
Push_Num(Num,(int)a[i]);
}
if(a[i+1]=='\0'){
Pop_Sq(Oper,x);
Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);
}
else if(a[i]==')'){
while(jud=GetTop_Sq(Oper,jud)!='('){
Pop_Sq(Oper,x);
Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);
}
Pop_Sq(Oper,x);
Push_Num(Num,result);
}
else {
if(Priority(a[i],jud=GetTop_Sq(Oper,jud))=='>'){ //当前操作符优先级高
Push_Sq(Oper,a[i]);
}
else if(Priority(a[i],jud=GetTop_Sq(Oper,jud))=='<'||Priority(a[i],jud=GetTop_Sq(Oper,jud))=='='){
Pop_Sq(Oper,x);
Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);
Push_Num(Num,result);
Push_Sq(Oper,a[i]);
}
}
i++;
}
//Pop_Sq(Oper,x);
/*Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);*/
printf("最终结果为%d",result);
}
int main(){
char a[5];
for(int i=0;i<5;i++){
scanf("%c",a[i]);
}
Init_Storage(a);
return 0;
}
/*
题目二
表达式求值问题
给定一个算术表达式,对该表达式进行四则运算。
基本要求:
(1)以栈作为存储结构,一个栈存放操作符,一个栈存放操作数;
(2)按字符串形式输入表达式;
(3)最终结果放在操作数栈中并输出。
测试数据要求:
给定的表达式中,必须有包含两位数的数值。例如,给定的表达式为:(10+2)*6-12/3
*/
#include <stdio.h>
#include <stdlib.h>
#define STACK_INIT_SIZE 100
#define STACKINCREMENT 30
typedef struct{
char *base;
int top;
int stacksize;
}SqStack;
typedef struct{
int *base;
int top;
int stacksize;
}NumStack;
void Init_SqStack(SqStack &S)
{
S.base = (char *)malloc(STACK_INIT_SIZE*sizeof(char));
if (!S.base) {
exit(0);
}
S.top = 0;
S.stacksize = STACK_INIT_SIZE;
}
int SqStackEmpty(SqStack S)
{
if(S.top == 0)
return 1;
else
return 0;
}
void Push_Sq(SqStack &S,char e)
{
if (S.top >= S.stacksize) {
S.base = (char *)realloc(S.base,(S.stacksize+STACKINCREMENT)*sizeof(char));
if(!S.base)
exit(0);
S.stacksize += STACKINCREMENT;
}
S.base[S.top++] = e;
}
void Pop_Sq(SqStack &S,char &e)
{
if(S.top==0){
exit(0);
}
e = S.base[--S.top];
}
char GetTop_Sq(SqStack S,char &e)
{
if(S.top == 0)
return -1;
e = S.base[S.top-1];
return e;
}
void Init_NumStack(NumStack &S)
{
S.base=(int *)malloc(STACK_INIT_SIZE*sizeof(int));
if(!S.base)
exit(0);
S.top = 0;
S.stacksize = STACK_INIT_SIZE;
}
int NumStackEmpty(NumStack S)
{
if(S.top==0)
return 1;
else
return 0;
}
void Push_Num(NumStack &S,int e)
{
if(S.top>=S.stacksize){
S.base=(int *)realloc(S.base,(S.stacksize+STACKINCREMENT)*sizeof(int));
if(!S.base){
exit(0);
}
S.stacksize+=STACKINCREMENT;
}
S.base[S.top++]=e;
return ;
}
void Pop_Num(NumStack &S,int &e){
if(S.top==0)
return;
e = S.base[--S.top];
}
void GetTop_Num(NumStack S,int &e)
{
if(S.top==0){
return ;
}
e=S.base[S.top-1];
return ;
}
char Priority(char a,char b)
{
if(a=='*'||a=='/'){
if(b=='+'||b=='-'){
return '>';
}else if(b=='*'||b=='/'){
return '=';
}else{
return '<';
}
}
if(a=='+'||a=='-'){
if(b=='+'||b=='-'){
return '=';
}
else{
return '<';
}
}
}
int calculate(int num1,char x,int num2)
{
if(x=='+'){
return num1+num2;
}
else if(x=='-'){
return num1-num2;
}
else if(x=='*'){
return num1*num2;
}
else if(x=='/'){
return num1/num2;
}
}
void Init_Storage(char a[])
{
NumStack Num;
SqStack Oper;
Init_SqStack(Oper);
Init_NumStack(Num);
Push_Sq(Oper,'#');
int i=0;
char x,jud;
int num1,num2,result=0;
while (a[i]) {
if(a[i]>='0'&&a[i]<='9')
Push_Num(Num,a[i] - '0'); /*不能直接强制类型转换*/
if(a[i+1] == 0){
Pop_Sq(Oper,x);
Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);
}
else if(a[i]==')') {
while((jud=GetTop_Sq(Oper,jud))!='('){
Pop_Sq(Oper,x);
Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);
}
Pop_Sq(Oper,x);
Push_Num(Num,result);
}
else {
if(Priority(a[i],jud=GetTop_Sq(Oper,jud))=='>'){ //当前操作符优先级高
Push_Sq(Oper,a[i]);
}
else if(Priority(a[i],jud=GetTop_Sq(Oper,jud))=='<'||Priority(a[i],jud = GetTop_Sq(Oper,jud))=='='){
Pop_Sq(Oper,x);
Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);
Push_Num(Num,result);
Push_Sq(Oper,a[i]);
}
}
i++;
}
//Pop_Sq(Oper,x);
/*Pop_Num(Num,num1);
Pop_Num(Num,num2);
result=calculate(num1,x,num2);*/
printf("最终结果为%d",result);
}
int main()
{
char a[80];
scanf("%s", a);
Init_Storage(a);
return 0;
}
参考一下吧
注意若a定义成5,很容易越界。
程序很接近答案了,建议你在这个基础上继续调试for(int i=0;i<5;i++){
scanf("%c", &a[i]);
}
少个&
/*---------------------------------------
函数型计算器(VC++6.0,Win32 Console)
功能:
目前提供了10多个常用数学函数:
⑴正弦sin
⑵余弦cos
⑶正切tan
⑷开平方sqrt
⑸反正弦arcsin
⑹反余弦arccos
⑺反正切arctan
⑻常用对数lg
⑼自然对数ln
⑽e指数exp
⑾乘幂函数^
⑿向上取整ceil
⒀向下取整floor
⒁四舍五入取整round
用法:
如果要求2的32次幂,可以打入2^32<回车>
如果要求30度角的正切可键入tan(Pi/6)<回车>
注意不能打入:tan(30)<Enter>
如果要求1.23弧度的正弦,有几种方法都有效:
sin(1.23)<Enter>
sin 1.23 <Enter>
sin1.23 <Enter>
如果验证正余弦的平方和公式,可打入sin(1.23)^2+cos(1.23)^2 <Enter>或sin1.23^2+cos1.23^2 <Enter>
此外两函数表达式连在一起,自动理解为相乘如:sin1.23cos0.77+cos1.23sin0.77就等价于sin(1.23)*cos(0.77)+cos(1.23)*sin(0.77)
当然你还可以依据三角变换,再用sin(1.23+0.77)也即sin2验证一下。
本计算器充分考虑了运算符的优先级因此诸如:2+3*4^2 实际上相当于:2+(3*(4*4))
另外函数名前面如果是数字,那么自动认为二者相乘.
同理,如果某数的右侧是左括号,则自动认为该数与括弧项之间隐含一乘号。
如:3sin1.2^2+5cos2.1^2 相当于3*sin2(1.2)+5*cos2(2.1)
又如:4(3-2(sqrt5-1)+ln2)+lg5 相当于4*(3-2*(√5 -1)+loge(2))+log10(5)
此外,本计算器提供了圆周率Pi键入字母时不区分大小写,以方便使用。
16进制整数以0x或0X开头。
----------------------------------------*/
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cmath>
#include <stdio.h>
#include <string.h>
#include <windows.h>
using namespace std;
const char Tab=0x9;
const int DIGIT=1;
const int MAXLEN=16384;
char s[MAXLEN],*endss;
int pcs=15;
double round(double dVal, short iPlaces) {//iPlaces>=0
char s[30];
double dRetval;
sprintf(s,"%.*lf",iPlaces,dVal);
sscanf(s,"%lf",&dRetval);
return (dRetval);
}
double fun(double x,char op[],int *iop) {
while (op[*iop-1]<32) //本行使得函数嵌套调用时不必加括号,如 arc sin(sin(1.234)) 只需键入arc sin sin 1.234<Enter>
switch (op[*iop-1]) {
case 7: x=sin(x); (*iop)--;break;
case 8: x=cos(x); (*iop)--;break;
case 9: x=tan(x); (*iop)--;break;
case 10: x=sqrt(x); (*iop)--;break;
case 11: x=asin(x); (*iop)--;break;
case 12: x=acos(x); (*iop)--;break;
case 13: x=atan(x); (*iop)--;break;
case 14: x=log10(x); (*iop)--;break;
case 15: x=log(x); (*iop)--;break;
case 16: x=exp(x); (*iop)--;break;
case 17: x=ceil(x); (*iop)--;break;
case 18: x=floor(x); (*iop)--;break;
case 19: x=round(x,0);(*iop)--;break;
}
return x;
}
double calc(char *expr,char **addr) {
static int deep; //递归深度
static char *fname[]={"sin","cos","tan","sqrt","arcsin","arccos","arctan","lg","ln","exp","ceil","floor","round",NULL};
double ST[10]={0.0}; //数字栈
char op[10]={'+'}; //运算符栈
char c,*rexp,*pp,*pf;
int ist=1,iop=1,last,i,n;
__int64 i64;
if (!deep) {
pp=pf=expr;
do {
c = *pp++;
if (c!=' '&& c!=Tab)
*pf++ = c;
} while (c!='\0');
}
pp=expr;
if ((c=*pp)=='-'||c=='+') {
op[0] = c;
pp++;
}
last = !DIGIT;
while ((c=*pp)!='\0') {
if (c=='(') {//左圆括弧
deep++;
ST[ist++]=calc(++pp,addr);
deep--;
ST[ist-1]=fun(ST[ist-1],op,&iop);
pp = *addr;
last = DIGIT;
if (*pp == '('||isalpha(*pp) && strnicmp(pp,"Pi",2)) {//目的是:当右圆括弧的右恻为左圆括弧或函数名字时,默认其为乘法
op[iop++]='*';
last = !DIGIT;
c = op[--iop];
goto operate ;
}
}
else if (c==')') {//右圆括弧
pp++;
break;
} else if (isalpha(c)) {
if (!strnicmp(pp,"Pi",2)) {
if (last==DIGIT) {
cout<< "π左侧遇)" <<endl;exit(1);
}
ST[ist++]=3.14159265358979323846264338328;
ST[ist-1]=fun(ST[ist-1],op,&iop);
pp += 2;
last = DIGIT;
if (!strnicmp(pp,"Pi",2)) {
cout<< "两个π相连" <<endl;exit(2);
}
if (*pp=='(') {
cout<< "π右侧遇(" <<endl;exit(3);
}
} else {
for (i=0; (pf=fname[i])!=NULL; i++)
if (!strnicmp(pp,pf,strlen(pf))) break;
if (pf!=NULL) {
op[iop++] = 07+i;
pp += strlen(pf);
} else {
cout<< "陌生函数名" <<endl;exit(4);
}
}
} else if (c=='+'||c=='-'||c=='*'||c=='/'||c=='%'||c=='^') {
char cc;
if (last != DIGIT) {
cout<< "运算符粘连" <<endl;exit(5);
}
pp++;
if (c=='+'||c=='-') {
do {
cc = op[--iop];
--ist;
switch (cc) {
case '+': ST[ist-1] += ST[ist];break;
case '-': ST[ist-1] -= ST[ist];break;
case '*': ST[ist-1] *= ST[ist];break;
case '/': ST[ist-1] /= ST[ist];break;
case '%': ST[ist-1] = fmod(ST[ist-1],ST[ist]);break;
case '^': ST[ist-1] = pow(ST[ist-1],ST[ist]);break;
}
} while (iop);
op[iop++] = c;
} else if (c=='*'||c=='/'||c=='%') {
operate: cc = op[iop-1];
if (cc=='+'||cc=='-') {
op[iop++] = c;
} else {
--ist;
op[iop-1] = c;
switch (cc) {
case '*': ST[ist-1] *= ST[ist];break;
case '/': ST[ist-1] /= ST[ist];break;
case '%': ST[ist-1] = fmod(ST[ist-1],ST[ist]);break;
case '^': ST[ist-1] = pow(ST[ist-1],ST[ist]);break;
}
}
} else {
cc = op[iop-1];
if (cc=='^') {
cout<< "乘幂符连用" <<endl;exit(6);
}
op[iop++] = c;
}
last = !DIGIT;
} else {
if (last == DIGIT) {
cout<< "两数字粘连" <<endl;exit(7);
}
if (pp[0]=='0' && (pp[1]=='x'||pp[1]=='X')) {
sscanf(pp+2,"%I64x%n",&i64,&n);
rexp=pp+2+n;
ST[ist++]=(double)i64;
} else ST[ist++]=strtod(pp,&rexp);
ST[ist-1]=fun(ST[ist-1],op,&iop);
if (pp == rexp) {
cout<< "非法字符" <<endl;exit(8);
}
pp = rexp;
last = DIGIT;
if (*pp == '('||isalpha(*pp)) {
op[iop++]='*';
last = !DIGIT;
c = op[--iop];
goto operate ;
}
}
}
*addr=pp;
if (iop>=ist) {
cout<< "表达式有误" <<endl;exit(9);
}
while (iop) {
--ist;
switch (op[--iop]) {
case '+': ST[ist-1] += ST[ist];break;
case '-': ST[ist-1] -= ST[ist];break;
case '*': ST[ist-1] *= ST[ist];break;
case '/': ST[ist-1] /= ST[ist];break;
case '%': ST[ist-1] = fmod(ST[ist-1],ST[ist]);break;
case '^': ST[ist-1] = pow(ST[ist-1],ST[ist]);break;
}
}
return ST[0];
}
int main(int argc,char **argv) {
int a;
if (argc<2) {
if (GetConsoleOutputCP()!=936) system("chcp 936>NUL");//中文代码页
cout << "计算函数表达式的值。"<<endl<<"支持(),+,-,*,/,%,^,Pi,sin,cos,tan,sqrt,arcsin,arccos,arctan,lg,ln,exp,ceil,floor,round"<<endl;
while (1) {
cout << "请输入表达式:";
gets(s);
if (s[0]==0) break;//
cout << s <<"=";
cout << setprecision(15) << calc(s,&endss) << endl;
}
} else if (argc==2 && 0==strcmp(argv[1],"/?")) {
if (GetConsoleOutputCP()!=936) system("chcp 936>NUL");//中文代码页
cout << "计算由≥1个命令行参数给出的函数表达式的值。\n"
"最后一个参数是.0~.15表示将计算结果保留小数0~15位\n"
"最后一个参数是x表示将计算结果以16进制正整数格式输出\n"
"支持(),+,-,*,/,%,^^,Pi,sin,cos,tan,sqrt,arcsin,arccos,arctan,lg,ln,exp,ceil,floor,round\n"
"16进制整数以0x或0X开头\n";
} else {
strncpy(s,argv[1],MAXLEN-1);s[MAXLEN-1]=0;
if (argc>2) {
for (a=2;a<argc-1;a++) strncat(s,argv[a],MAXLEN-1);//将空格间隔的各参数连接到s
if (1==sscanf(argv[a],".%d",&pcs) && 0<=pcs && pcs<=15) {//最后一个参数是.0~.15表示将计算结果保留小数0~15位
printf("%.*lf\n",pcs,calc(s,&endss));
} else if (argv[a][0]=='x'||argv[a][0]=='X') {//最后一个参数是x表示将计算结果以16进制正整数格式输出
printf("0x%016I64x\n",(__int64)calc(s,&endss));
} else {
strncat(s,argv[a],MAXLEN-1);
printf("%.15lg\n",calc(s,&endss));
}
} else {
printf("%.15lg\n",calc(s,&endss));
}
}
return 0;
}