UVA 327 Evaluating Simple C Expressions

UVA_327

    類比題目所說的即可。

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#define MAXD 200
int pre[MAXD], vis[MAXD], last[MAXD];
char b[MAXD], st1[MAXD], st2[MAXD], st3[MAXD];
void solve()
{
int i, j, k, n1, n2, n3, res;
for(i = 'a'; i <= 'z'; i ++)
    {
        pre[i] = i - 'a' + 1;
        last[i] = vis[i] = 0;
    }
for(i = n1 = 0; b[i]; i ++)
    {
if(isalpha(b[i]))
            vis[b[i]] = 1;
if(b[i] != ' ')
            st1[n1 ++] = b[i];
    }
for(i = n2 = 0; i < n1; i ++)
    {
if(i + 2 < n1 && st1[i] == '+' && st1[i + 1] == '+' && isalpha(st1[i + 2]))
        {
            ++ pre[st1[i + 2]];
            ++ i;
        }
else if(i + 2 < n1 && st1[i] == '-' && st1[i + 1] == '-' && isalpha(st1[i + 2]))
        {
            -- pre[st1[i + 2]];
            ++ i;
        }
else
            st2[n2 ++] = st1[i];
    }
for(i = n3 = 0; i < n2; i ++)
    {
        st3[n3 ++] = st2[i];
if(i + 2 < n2 && isalpha(st2[i]) && st2[i + 1] == '+' && st2[i + 2] == '+')
        {
            ++ last[st2[i]];
            i += 2;
        }
else if(i + 2 < n2 && isalpha(st2[i]) && st2[i + 1] == '-' && st2[i + 2] == '-')
        {
            -- last[st2[i]];
            i += 2;
        }
    }
    res = pre[st3[0]];
for(i = 1; i < n3; i += 2)
    {
if(st3[i] == '+')
            res += pre[st3[i + 1]];
else
            res -= pre[st3[i + 1]];
    }
    printf("    value = %d\n", res);
for(i = 'a'; i <= 'z'; i ++)
if(vis[i])
            printf("    %c = %d\n", i, pre[i] + last[i]);
}
int main()
{
while(gets(b) != NULL)
    {
        printf("Expression: %s\n", b);
        solve();
    }
return 0;
}