【ZZNU-oj-2116:人间不值得】（1亿以内的货币拼音转数值求折扣价原价）--hash+String大法好+字符串处理+超大暴力模拟题

  1 #include <iostream>
  2 #include<stdio.h>
  3 #include<string.h>
  4 #include<string>
  5 #include<algorithm>
  6 #define ll long long
  7 using namespace std;
  8 #define N 100
  9 char arr[20][10]= {"ling","yi","er","san","si","wu","liu","Qi","Ba","jiu",
 10                    "shi","bAi","qIan","wan"
 11                   };  //0~9 + 10~13
 12 int has[]= {0,1,2,3,4,5,6,7,8,9,10,100,1000,10000,100000000}; //has[14]为亿
 13 int d[]= {4,2,2,3,2, 2,3,2,2,3, 3,3,4,3}; //偏移量
 14 int solve(char sn[])
 15 {
 16 
 17     string s(sn);
 18     int x,len=s.length();
 19 
 20 
 21     while(x=s.find("qi"),x>=0&&x<len)  //替换qi 和 qian
 22     {
 23         if(x+2<=len-1&&s[x+2]!=‘a‘)
 24             s[x]=‘Q‘;
 25         else if(x+2>=len)
 26             s[x]=‘Q‘;
 27         else
 28             s[x+1]=‘I‘;
 29     }
 30     while(x=s.find("ba"),x>=0&&x<len)  //替换Ba 和 bAi
 31     {
 32         if(x+2<=len-1&&s[x+2]!=‘i‘)
 33             s[x]=‘B‘;
 34         else if(x+2>=len)
 35             s[x]=‘B‘;
 36         else
 37             s[x+1]=‘A‘;
 38     }
 39     x=s.find("yuan");  //删掉末尾的 yuan
 40     if(x>=0&&x<len)
 41         s.erase(x,4);
 42     // cout<<s<<" ~~删减yuan后~"<<endl;
 43     //特判1亿元
 44     x=s.find("yiyi");
 45     if(x>=0&&x<len)
 46         return has[14];
 47 
 48     x=0;
 49     len=s.length();
 50     int sum=0,ss=0;
 51     while(x<len)
 52     {
 53         int i=0;//代表是第几个汉字
 54         while(i<14&&s.find(arr[i])!=x)
 55             i++;
 56         if(i==14)     //出现错误，反馈error
 57         {
 58             cout<<"error"<<endl;
 59             break;
 60         }
 61         if(i>=10&&i<=12)
 62         {
 63             if(ss==0)sum+=has[i];
 64             ss*=has[i],sum+=ss,ss=0;
 65         }
 66         else if(i==13)
 67             sum+=ss,ss=0,sum*=has[13];
 68         else if(i>0)
 69             ss=has[i];
 70 
 71         for(int j=x; j<x+d[i]; j++) //走一步删一步
 72             s[j]=‘*‘;
 73         x+=d[i];
 74         // printf("***i=%d x=%d ss=%d sum=%d\n",i,x,ss,sum);
 75     }
 76     return sum+ss;
 77 }
 78 double get_op(char sn[])
 79 {
 80     string s(sn);
 81     int x,len=s.length();
 82     double op=0.0;
 83     while(x=s.find("qi"),x>=0&&x<len)  //替换qi 和 qian
 84     {
 85         if(x+2<=len-1&&s[x+2]!=‘a‘)
 86             s[x]=‘Q‘;
 87         else if(x+2>=len)
 88             s[x]=‘Q‘;
 89         else
 90             s[x+1]=‘I‘;
 91     }
 92     while(x=s.find("ba"),x>=0&&x<len)  //替换Ba 和 bAi
 93     {
 94         if(x+2<=len-1&&s[x+2]!=‘i‘)
 95             s[x]=‘B‘;
 96         else if(x+2>=len)
 97             s[x]=‘B‘;
 98         else
 99             s[x+1]=‘A‘;
100     }
101     int sum=0,ss=0;
102     x=s.find("zhe");  //删掉末尾的 zhe
103     if(x>=0&&x<len)
104         s.erase(x,4);
105     len=s.length();
106     x=0;
107     int cnt=0;
108     while(x<len)
109     {
110         int i=0;//代表是第几个个汉字
111         while(i<14&&s.find(arr[i])!=x)
112             i++;
113         if(i==14)
114         {
115             cout<<"error"<<endl;
116             break;
117         }
118         op=op*10.0+has[i];
119         cnt++;
120         for(int j=x; j<x+d[i]; j++) //走一步删一步
121             s[j]=‘*‘;
122         x+=d[i];
123         //   printf("***i=%d x=%d op=%lf \n",i,x,op);
124     }
125     while(cnt--)
126         op=op*0.1;
127     return op;
128 }
129 int main()
130 {
131     char s1[100],s2[100],s3[100];
132     while(scanf("%s%s%s",s1,s2,s3)!=EOF)
133     {
134         if(strcmp(s1,"yuanjia")==0)   //给的是原价和折扣op
135         {
136             double cost,discost,op;  //原价为cost，折扣后的价为discost！
137             cost=solve(s2)*1.0;
138             op=get_op(s3);//printf("op*****%.6lf\n",op);
139 
140             discost=cost*op;
141             printf("%.2lf %.2lf\n",cost,discost);
142         }
143         else   //给的是折扣价和 折扣op
144         {
145             double cost,discost,op;
146             discost=solve(s2);
147             op=get_op(s3);//printf("op*****%.6lf\n",op);
148             cost=discost/op;
149             printf("%.2lf %.2lf\n",cost,discost);
150         }
151         //  printf("****%.4lf\n",1.11/10.00);
152     }
153 
154     return 0;
155 }