信息学奥赛一本通第三部分_栈 ex1_4cale （中缀转后缀7符号）

Posted 2020-06-14

　　其实这个中缀转后缀是费了很大功夫的，明白算法后第一次实现花了近三小时ORZ

#include <stdio.h>
#include <string.h>
#include <ctype.h>

char Mstr[511],Msta[511] = {‘@‘},Bstr[511];
int sta[511];
const short list[4][4] = {{0,-1,1,1},{1,0,1,1},{1,1,1,-2},{-1,-1,2,1}};

int level (char c){
    switch (c){
        case ‘+‘:
        case ‘-‘:return 0;break;
        case ‘*‘:
        case ‘/‘:return 1;break;
        case ‘(‘:return 2;break;
        case ‘)‘:return 3;
    }
}

int main (){
//    freopen ("中后6.in","r",stdin);
//    freopen (".out","w",stdout);
    
    _Bool flag = 1;
    int i,Br = -1,Ma = 0;
    
    gets (Mstr);
    for (i = 0;Mstr[i-1]!=‘@‘;i++){
        if (isdigit(Mstr[i])){
            Bstr[++Br] = Mstr[i];
            flag = 1;
        }
        else{
            if (flag == 1){
                Bstr[++Br] = ‘ ‘;
                flag = 0;
            }
            int lr = level(Mstr[i]),la = level(Msta[Ma]);
            int res;
            if (Msta[Ma] == ‘@‘)
                res = 1;
            else if (Mstr[i] == ‘@‘)
                res = -1;
            else
                res = list[lr][la];
            
            if (res == 2){
                Ma--;
                continue;
            }
            do{
                switch (res){
                    case 1:Msta[++Ma] = Mstr[i];break;
                    case -1:Bstr[++Br] = Msta[Ma--];res = list[lr][level(Msta[Ma])];break;
                    case -2:printf ("input error!");return 0;
                }
                if (res == 2)
                    Ma--;
            }while (res == -1);
            if (res == 0){
                Bstr[++Br] = Msta[Ma];
                Msta[Ma] = Mstr[i];
            }
        }
    }
    
    printf ("%s",Bstr);

    return 0;
}

初版6符号

　　但还是逻辑性不强，而且感觉还是没理解透，于是又作死花了近两小时隔天打了一遍。用长变量名达到不需要注释，逻辑性自认增强很多

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <ctype.h>
  4 
  5 char Mstr[32] = {0},Msta[32] = {0},Bstr[32] = {0};
  6 int Bsta[32] = {0};
  7 int MstrLen,BstrLen,MstaTop,BstrTop,BstaTop;
  8 
  9 const short kind (char c){
 10     switch (c){
 11         case ‘)‘: return -1;
 12         case ‘@‘: return 0;
 13         case ‘+‘:
 14         case ‘-‘: return 1;
 15         case ‘*‘: 
 16         case ‘/‘: return 2;    
 17         case ‘(‘: return 3;
 18         case ‘^‘: return 4;
 19     }
 20 }
 21 
 22 void Mstr_Bsta (void){
 23     gets (Mstr); 
 24     MstrLen = strlen (Mstr); 
 25     Mstr[MstrLen++] = ‘@‘;
 26     MstaTop = BstrTop = -1;
 27     Msta[++MstaTop] = ‘@‘;
 28     
 29     int SpaceFlag,NoSmallerFlag;
 30     int i;
 31     
 32     for (i = 0;i<MstrLen;i++){
 33         if (isdigit(Mstr[i])){
 34             Bstr[++BstrTop] = Mstr[i];
 35             SpaceFlag = 1;
 36         }
 37         else{
 38             if (SpaceFlag == 1){
 39                 Bstr[++BstrTop] = ‘ ‘;
 40                 SpaceFlag = 0;
 41             }
 42             
 43             int KindMstr,KindMsta;
 44             do{
 45                 NoSmallerFlag = 1;
 46                 KindMstr = kind(Mstr[i]); 
 47                 KindMsta = kind(Msta[MstaTop]);
 48                 
 49                 if (KindMstr>KindMsta || (KindMstr == 3 && KindMsta == 4) || (KindMsta == 3 && KindMstr!=-1))
 50                     Msta[++MstaTop] = Mstr[i];
 51                 else if (KindMstr == KindMsta){
 52                     Bstr[++BstrTop] = Msta[MstaTop];
 53                     Msta[MstaTop] = Mstr[i];
 54                 }
 55                 else{
 56                     if (KindMstr == -1 && KindMsta == 3){
 57                         --MstaTop;
 58                         break;
 59                     }
 60                     else
 61                         Bstr[++BstrTop] = Msta[MstaTop--];
 62                     NoSmallerFlag = 0;    
 63                 }
 64             }while (KindMstr<KindMsta && NoSmallerFlag == 0);
 65         }
 66     }
 67 }
 68 
 69 int PowerQuickly (int x,int n){    //快速幂 
 70     int ans = 1;
 71     
 72     while (n>0){
 73         if (n&1 == 1)
 74             ans*=x;
 75         n/=2;
 76         x*=x;    
 77     }
 78     return ans;
 79 }
 80 
 81 void Bsta_Count (void){
 82     int i;
 83     
 84     for (i = 0,BstaTop = 0;Bstr[i]!=‘@‘;){
 85         if (isdigit(Bstr[i])){
 86             int num = Bstr[i++]-‘0‘;
 87             while (Bstr[i]!=‘ ‘)
 88                 num = num*=10+Bstr[i++]-‘0‘;
 89             Bsta[++BstaTop] = num;
 90             i++;    
 91         }
 92         else{
 93             switch (Bstr[i++]){
 94                 case ‘+‘: Bsta[BstaTop-1] += Bsta[BstaTop--];break;
 95                 case ‘-‘: Bsta[BstaTop-1] -= Bsta[BstaTop--];break;
 96                 case ‘*‘: Bsta[BstaTop-1] *= Bsta[BstaTop--];break;
 97                 case ‘/‘: Bsta[BstaTop-1] /= Bsta[BstaTop--];break;
 98                 case ‘^‘: Bsta[BstaTop-1] = PowerQuickly(Bsta[BstaTop-1],Bsta[BstaTop]); BstaTop--;
 99             }
100         }
101     }
102 }
103 
104 int main (){    
105     Mstr_Bsta ();
106     Bsta_Count ();
107     printf ("%d",Bsta[BstaTop]);
108     
109     return 0;
110 }

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