用c程式設計 可混合運算的計算器
阿新 • • 發佈：2019-01-06
功能
有優先順序的運算
八進位制與十六進位制
十六進位制與八進位制
有優先順序的運算
八進位制與十六進位制
十六進位制與八進位制
一元多項式求和
#include <stdio.h> #include <stdlib.h> #include <math.h> #include <string.h> #define STACK_INIT_SIZE 20 #define STACKINCREMENT 10 char OP[]={'+','-','*','/','(',')','='}; typedef char ElemType; struct node { int coef; //係數 int exp; //指數 struct node * next;//指標域 }chainLink;//建立chainLink的node結點物件 typedef struct { ElemType *base; ElemType *top; int stackSize; }sqStack; typedef struct { int *base; int *top; int stackSize; }SqStack; void InitStack(sqStack *s) { s->base = (ElemType *)malloc(STACK_INIT_SIZE * sizeof(ElemType)); if( !s->base ) { exit(0); } s->top = s->base; s->stackSize = STACK_INIT_SIZE; } void initStack(SqStack *s) { s->base = (int *)malloc(STACK_INIT_SIZE * sizeof(int)); if( !s->base ) { exit(0); } s->top = s->base; s->stackSize = STACK_INIT_SIZE; } void Push(sqStack *s, ElemType e) { if( s->top - s->base >= s->stackSize ) { s->base = (ElemType *)realloc(s->base, (s->stackSize + STACKINCREMENT) * sizeof(ElemType)); if( !s->base ) { exit(0); } } *(s->top) = e; s->top++; } void Pop(sqStack *s, ElemType *e) { if( s->top == s->base ) { return; } *e = *--(s->top); } void 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(0); } } *(s->top) = e; s->top++; } void pop(SqStack *s, int *e) { if( s->top == s->base ) { return; } *e = *--(s->top); } int StackLen(sqStack s) { return (s.top - s.base); } int stackLen(SqStack s) { return (s.top - s.base); } char GetTop(sqStack *s) { char e; if(s->top == s->base) return -1; e = *(s->top-1); return e; } int getTop(SqStack *s) { int e; if(s->top == s->base) return -1; e = *(s->top-1); return e; } int judge(char theta) { int i; if(theta == '+') i=0; else if(theta == '-') i=1; else if(theta == '*') i=2; else if(theta == '/') i=3; else if(theta == '(') i=4; else if(theta == ')') i=5; else i=6; return i; } int precede(char theta1,char theta2) { int std_cherry[7][7]= { 1, 1,-1,-1,-1, 1, 1, 1, 1,-1,-1,-1, 1, 1, 1, 1, 1, 1,-1, 1, 1, 1, 1, 1, 1,-1, 1, 1, -1,-1,-1,-1,-1, 0,-2, 1, 1, 1, 1,-2, 1, 1, -1,-1,-1,-1,-1,-2, 0, }; int i,j; i = judge(theta1); j = judge(theta2); return std_cherry[i][j]; } int OPerate(int a,char theta,int b) { int result; switch(theta) { case '+': result = a+b; break; case '-': result = a-b; break; case '*': result = a*b; break; case '/': result = a/b; break; } return result; } int In(char *c,char *OP) { int i,comp = 0; for(i = 0; i < 7; i++) { if( c[0] == OP[i]) comp = 1; } return comp; } int EvaluateEcpression() { sqStack OPTR; SqStack OPND; char c[10],theta,x; int a,b,put; InitStack(&OPTR); Push(&OPTR,'='); initStack(&OPND); scanf("%s",c); while(c[0] != '=' || GetTop(&OPTR) != '=') { if(!In(c,OP)) { put = atoi(c); // printf("%d\n",put); push(&OPND,put); scanf("%s",c); } else switch(precede(GetTop(&OPTR),c[0])) { case -1: Push(&OPTR,c[0]); scanf("%s",c); break; case 0: Pop(&OPTR,&x); scanf("%s",c); break; case 1: Pop(&OPTR,&theta); pop(&OPND,&b); pop(&OPND,&a); push(&OPND,OPerate(a,theta,b)); // printf("%d\n",OPerate(a,theta,b)); break; } } return getTop(&OPND); } void BtoD() { ElemType c; sqStack s; int len, i, sum = 0; InitStack(&s); printf("請輸入二進位制數，輸入#符號表示結束！\n"); scanf("%c", &c); while( c != '#' ) { Push(&s, c); scanf("%c", &c); } getchar(); // 把'\n'從緩衝區去掉 len = StackLen(s); printf("棧的當前容量是: %d\n", len); for( i=0; i < len-1; i++ ) { Pop(&s, &c); sum = sum + (c-48) * pow(2, i); } printf("轉化為十進位制數是: %d\n", sum); return 0; } void DtoB() { int c,d; SqStack s; int len, i; InitStack(&s); printf("請輸入十進位制數: "); scanf("%d", &c); printf("轉化為二進位制數是: "); while(c > 0) { // printf("%d",c%2); push(&s, c%2); c = c - c%2; c = c/2; } len = stackLen(s); for(i = 0; i < len; i++) { pop(&s,&d); printf("%d",d); } printf("\n棧的當前容量是: %d\n", len); return 0; } void OtoH() { int i,j,k,s=0; int a; char p[30]; int result=0; printf("輸入你想要轉化的八進位制數: \n"); scanf("%d",&a); for(i=1;a!=0;i*=8) { if(a%10>7) { s=1; break; } else { result+=(a%10)*i; a=a/10; } } for(j=0;result!=0;j++) { p[j]=result%16; result=result/16; if(p[j]<10) p[j]+=48; else { switch(p[j]) { case 10: p[j]='A'; break; case 11: p[j]='B'; break; case 12: p[j]='C'; break; case 13: p[j]='D'; break; case 14: p[j]='E'; break; case 15: p[j]='F'; break; } } } if(s==1) printf("您的輸入有誤！請重新輸入\n"); else { printf("\n轉換後的數為："); for(k=j-1;k>=0;k--) { printf("%c",p[k]); } printf("\n"); } } void HtoO() { int i,j,k,s=0; char a[10]; int result=0; int b[30]; int p[30]; scanf("%s",&a); for(k=0;;k++) { if(a[k]=='\0') break; } for(i=0;i<k;i++) { if(a[i]<='9'&&a[i]>='1') b[i]=a[i]-48; else { switch(a[i]) { case 'A': b[i]=10; break; case 'B': b[i]=11; break; case 'C': b[i]=12; break; case 'D': b[i]=13; break; case 'E': b[i]=14; break; case 'F': b[i]=15; break; case 'a': b[i]=10; break; case 'b': b[i]=11; break; case 'c': b[i]=12; break; case 'd': b[i]=13; break; case 'e': b[i]=14; break; case 'f': b[i]=15; break; default: s=1; } } } for(j=k-1,i=1;j>=0;j--,i*=16) { result+=b[j]*i; } for(j=0;result!=0;j++) { p[j]=result%8; result=result/8; } if(s==1) printf("您的輸入有誤！請重新輸入\n"); else { printf("\n轉換後的數為："); for(k=j-1;k>=0;k--) { printf("%d",p[k]); } printf("\n"); } } struct node *create() //定義建立多項式函式，並返回連結串列頭指標 { struct node *h = NULL, *q, *p;//定義結構體頭指標h，標記指標p和q，p是創造新節點的標記指標，q是連結連結串列的指標； int i = 1, N; //定義多項式的項數N及標記數i printf("請輸入多項式的項數: "); scanf("%d",&N); p = 0;//指標初始化； q = 0;//本地指標初始化； for (; i <= N; i++) { p = (struct node *)malloc(sizeof(chainLink)); //為一個新節點分配空間 printf("請輸入第 %i 項的係數：\t",i); scanf("%d",&(*p).coef); printf("請輸入第 %i 項的指數: \t",i); scanf("%d",&(*p).exp); if (i == 1) { h = p; q = p; } //建立頭節點 else { q->next = p; q = p; } } q->next = NULL;//p,q都成為新連結串列的尾指標； p->next = NULL; return h; } void display(struct node *h) { struct node *p; //定義標記指標，輸出連結串列 p = h; while (p != NULL) { if (p->coef == 0) { struct node *d; d = h; while (d->next != p) { d = d->next; } d->next = p->next; p = p->next;//刪去係數為0的節點； } else { if (p->coef<0) //係數為負時應加上括號,讓式子容易看清； { if (p->exp == 0) printf("%d+",(*p).coef); else printf("(%d)*X^%d+",(*p).coef,(*p).exp); } else { if (p->exp == 0) printf("%d+",(*p).coef); else if (p->exp!=0&&p->exp!=1) { printf("%d*X^%d+",(*p).coef,(*p).exp); } else printf("%d*X+",(*p).coef); } p = p->next; } } printf("\b \b \n"); } struct node *order(struct node *h) //定義排序函式，並返回整理後的連結串列的頭指標 { struct node *p, *q, *t, *h1, *h2; //定義三個結構體標記指標和兩個頭指標 h1 = h; //建立一個新的連結串列，頭指標為h，指向原連結串列的第一個節點，之後將原連結串列中的節點一個一個的插入此連結串列，進行排序 h2 = h1->next; //將原來的連結串列建立成待排序連結串列 h1->next = NULL; //截斷第一個原連結串列中的節點 while (h2 != NULL) { q = h1; p = q->next; t = h2; //從待排序連結串列中選出一個節點準備插入到新連結串列中 h2 = h2->next; //移動待排序連結串列的頭指標，便於進行下一次挑選 t->next = NULL; if (t->exp>q->exp) //應插入頭指標之前的情況 { t->next = q; h1 = t; continue; } if (p == NULL&&t->exp <= q->exp) q->next = t; //應插入表尾的情況 while (p != NULL) { if (t->exp>p->exp) { q->next = t; t->next = p; break; } else { q = q->next; p = p->next; } } if (p == NULL) q->next = t; } return h1; //新連結串列即為按降冪順序排好的連結串列，返回其頭指標 } struct node *add(struct node *h1, struct node *h2) //定義加法函式，並返回結果的連結串列的頭指標 { struct node *p, *q, *head, *r; //定義結構體頭指標head和標記指標p,q,r p = h1; q = h2; head = (struct node *)malloc(sizeof(chainLink)); //為結果多項式分配頭指標 if (p->exp >= q->exp) { head = h1; p = p->next; } else { if (p->exp<q->exp) { head = h2; q = q->next; } else { p->coef = p->coef + q->coef; head = p; p = p->next; q = q->next; } } r = head; while (p != NULL&&q != NULL) { if (p->exp>q->exp) { r->next = p; p = p->next; } else { if (p->exp<q->exp) { r->next = q; q = q->next; } else { p->coef = p->coef + q->coef; r->next = p; p = p->next; q = q->next; } } r = r->next; } if (p == NULL&&q == NULL) r->next = NULL; else { if (p == NULL) r->next = q; if (q == NULL) r->next = p; } return head; } void polynomail() { struct node *h1, *h2, *h3;//定義3個頭指標； printf("請輸入第一個多項式：\n"); h1 = create();//創造第一個線性連結串列； printf("\n您輸入的第一個多項式為：\n"); display(h1);//把多項式顯示出來； h1 = order(h1); printf("\n降冪排列後的第一個多項式為：\n"); display(h1); printf("\n"); printf("\n請輸入第二個多項式：\n"); h2 = create();//創造第二個線性連結串列； printf("\n您輸入的第二個多項式為：\n"); display(h2);//把多項式顯示出來； h2 = order(h2); printf("\n降冪排列後的第二個多項式為：\n"); display(h2); printf("\n"); h3 = add(h1, h2);//利用加函式將多項式相加； printf("\n第一個多項式和第二個多項式相加後：\n"); display(h3); system("pause"); return 0; } void phead(void) { printf("*******************************************************\n"); printf("********************Tom's calclater********************\n"); printf("*******************************************************\n"); printf("\n*************** 請選擇需要的計算操作:*****************\n"); } void thead(void) { printf("*******************************************************\n"); printf("請輸入相應的數字,並進行一步計算: "); } void main() { int i,j,q=0; int tom; phead(); do { printf("0、退出\n1、進位制轉化\n2、計算器\n3、一元多項式加法\n"); thead(); scanf("%d",&i); switch (i) { case 1: thead(); printf("\n請要轉化的的進位制:\n0、退出\n1、二進位制到十進位制\n2、十進位制到二進位制\n3、八進位制到十六進位制\n4、十六進位制到八進位制\n"); scanf("%d",&j); switch(j) { case 1: printf("\n請輸入您想要轉化的數："); BtoD(); break; case 2: printf("\n請輸入您想要轉化的數："); DtoB(); break; case 3: printf("\n請輸入您想要轉化的數："); OtoH(); break; case 4: printf("\n請輸入您想要轉化的數："); HtoO(); break; case 0: printf("謝謝使用！！"); } break; case 2: printf("\n請輸入你想要計算的表示式（在以空格隔開每個輸入！）：\n"); tom = EvaluateEcpression(); printf("表示式的計算結界是 %d\n",tom); break; case 3: polynomail(); break; case 0: printf("\n謝謝使用！\n"); break; } }while(q==1); }