The Chinese production of simple compiler

This work belongs courses	https://edu.cnblogs.com/campus/fzzcxy/SE
Where the job requires	https://edu.cnblogs.com/campus/fzzcxy/SE/homework/10283
The target job	Pick up again down the C language freshman of knowledge
Text of the job	as follows
Other references	Baidu various reference materials, as well as to gain knowledge of other heavyweights blog

Lines of code: 158 lines
needs analysis: 2hour
write code: 2 days

Problems encountered:
1, new to the C language, and some grammar forgotten
2, the meaning of the title is not comprehensive understanding of
3, clear thinking subject, however, you need to draw on the knowledge and Baidu blog big brother

code show as below:

include<stdio.h>

include<string.h>

include<math.h>

int first(char a[])
{
if(strcmp("零",a)==0)
return 0;
if(strcmp("一",a)==0)
return 1;
if(strcmp("二",a)==0)
return 2;
if(strcmp("三",a)==0)
return 3;
if(strcmp("四",a)==0)
return 4;
if(strcmp("五",a)==0)
return 5;
if(strcmp("六",a)==0)
return 6;
if(strcmp("七",a)==0)
return 7;
if(strcmp("八",a)==0)
return 8;
if(strcmp("九",a)==0)
return 9;
if(strcmp("十",a)==0)
return 10;
}
int second(char a[]){
char b[10],c[10];
if(strlen(a)==2)
{
return first(a);
}
else if(strlen(a)==4)
{
b[0]=a[2];
b[1]=a[3];
b[2]='\0';
if(first(b)!=10)
return 10+first(b);
else{
b[0]=a[0];
b[1]=a[1];
b[2]='\0';
return 10*first(b);
}

}
else if(strlen(a)==6)
{
    b[0]=a[4];
    b[1]=a[5];
    b[2]='\0';
    c[0]=a[0];
    c[1]=a[1];
    c[2]='\0';
    return first(b)+first(c)*10;
}

}
int main(){
char *chinese[11]={"零","一","二","三","四","五","六","七","八","九","十"};
char name[200],verb[200],numb[200],oper[200],name1[200],toint[200],name3[200];
char x[20],y[20],numb2[20],d[20],word[100],e[20],f[20],cold[100],name2[100];
char age[200],money[200];
int sum,sum2,sum3;
scanf("%s",toint);
scanf("%s %s %s",name,verb,numb);
sum=second(numb);
while(scanf("%s",name1)!=EOF)
{
if(strcmp(name1,toint)==0)
{
scanf("%s %s %s",name3,verb,numb);
sum2=first(numb);
}
if(strcmp(name1,name)==0){
scanf("%s",oper);
if(strcmp(oper,"增加")==0){
scanf("%s",numb);
sum3=second(numb);
sum+=sum3;}
if(strcmp(oper,"减少")==0){
scanf("%s",numb);
if(strlen(numb)==2)
{
sum3=second(numb);
sum-=sum3;
}
}
}
else if(strcmp(name1,"看看")==0)
{
int lasttmp;
scanf("%s",name);
scanf("%s %s %s %s %s %s %s %s %s %s",x,name,y,numb2,d,name2,word,e,f,cold);
lasttmp=first(numb2);
int len=strlen(word);
int len1=strlen(cold);
{
if(sum>=lasttmp)
{ if(sum>=0&&sum<=10){
printf("%s\n",chinese[sum]);for (int i=1;i<len-1; i+=2) {printf("%c%c",word[i],word[i+1]);}
}
else {
if(sum%10==0){
printf("%s十\n",chinese[sum/10]);for (int i=1;i<len-1; i+=2) {printf("%c%c",word[i],word[i+1]);}
}
else if(sum<=19){
printf("十%s\n",chinese[sum%10]);for (int i=1;i<len-1; i+=2) {printf("%c%c",word[i],word[i+1]);}
}
else {
printf("%s十%s\n",chinese[sum/10],chinese[sum%10]);for (int i=1;i<len-1; i+=2) {printf("%c%c",word[i],word[i+1]);}
}
}
}
else if(sum<lasttmp){
if(sum<0)
{
int under=abs(sum);
printf("%s%s\n","负",chinese[under]);
for(int j=1;j<len1-1;j+=2) {printf("%c%c",cold[j],cold[j+1]);}}
}
else{
printf("%s\n",chinese[sum]);for(int j=1;j<len1-1;j+=2) {printf("%c%c",cold[j],cold[j+1]);}}
}
}
else if(strcmp(name1,"如果")==0)
{
scanf("%s %s %s %s %s %s %s %s %s",age,y,numb2,d,money,word,e,f,cold);
scanf("%s %s",x,age);
if(strcmp(x,"看看")==0)
{
if(sum>8)
{
int x=sum2+1;
printf("%s",chinese[x]);
}
}
}
}
return 0;
}

Example 1:
integer temperature equal to ten
temperatures reduce the three
temperature increase of two
to see temperatures
If the temperature is greater than eight to see "Hello, world" or look at the "frozen to death of me."

Output:

Sample 2:
integer equal to eight age-Xiao Yang
integer Xiao Yang allowance equal to two
to increase a Xiao Yang of age
increased if Xiao Yang Xiao Yang older than eight or no pocket money a
look at Xiao Yang pocket money

Output: