327 Evaluating Simple C Expressions

描述：題目很簡單，剛開始考慮問題時忽略了所有的字母所代表的值需要全部變化後才進行運算，結果交了幾次錯誤碼，悲催啊……#include <iostream>#include <cstring>#include <cstdio>using namespace std;int main(){    //freopen("a.txt","r",stdin);    int n,i,j;    int s_flag[27],s_add[27],s_value[27];    int s_untake[10010];    char str[10010],s[10010];    memset(str,0,sizeof(str));    while(gets(str))    {        j=n=0;        printf("Expression: %s\n",str);        memset(s_flag,0,sizeof(s_flag));        memset(s_add,0,sizeof(s_add));        memset(s,0,sizeof(s));        memset(s_value,0,sizeof(s_value));        memset(s_untake,0,sizeof(s_untake));        for(i=0; i<strlen(str); i++)        {            if(str[i]>='a'&&str[i]<='z')            {                s_flag[str[i]-'a']=1;                s_value[str[i]-'a']=str[i]-'a'+1;            }            if((str[i]>='a'&&str[i]<='z')||str[i]=='-'||str[i]=='+')s[j++]=str[i];        }        memset(str,0,sizeof(str));        for(i=0; i<j; i++)            if(s[i]>='a'&&s[i]<='z')            {                if(i>=2)                {                    if(s[i-1]=='+'&&s[i-2]=='+')                    {                        s_value[s[i]-'a']++;                        s_untake[i-1]=s_untake[i-2]=1;                    }                    else if(s[i-1]=='-'&&s[i-2]=='-')                    {                        s_value[s[i]-'a']--;                        s_untake[i-1]=s_untake[i-2]=1;                    }                }                if(i<j-2)                {                    if(s[i+1]=='+'&&s[i+2]=='+')                    {                        s_add[s[i]-'a']++;                        s_untake[i+1]=s_untake[i+2]=1;                    }                    else if(s[i+1]=='-'&&s[i+2]=='-')                    {                        s_add[s[i]-'a']--;                        s_untake[i+1]=s_untake[i+2]=1;                    }                }            }        for(i=0; i<j; i++)            if(s[i]<='z'&&s[i]>='a')            {                n=s_value[s[i]-'a'];                i++;                break;            }        for(; i<j; i++)            if(s[i]>='a'&&s[i]<='z')            {                if(i-1>=0&&!s_untake[i-1])                {                    if(s[i-1]=='+') n+=s_value[s[i]-'a'];                    else if(s[i-1]=='-') n-=s_value[s[i]-'a'];                }                if(i-3>=0&&(s[i-2]=='+'||s[i-2]=='-')&&!s_untake[i-3])                {                    if(s[i-3]=='+') n+=s_value[s[i]-'a'];                    else if(s[i-3]=='-') n=n-s_value[s[i]-'a'];                }            }        printf("    value = %d\n",n);        for(i=0; i<26; i++)            if(s_flag[i]==1)printf("    %c = %d\n",'a'+i,s_add[i]+s_value[i]);    }    return 0;}