不开O_2 就能过

开O_2就RE

#include <bits/stdc++.h>
#define int long long
using namespace std;
const int maxn = 1010;
char c[maxn];
int st[maxn];
bool comp(char *c)
{
    int i = 0;
    int now = 0;
    int top = 0;
    while(c[i]!='@')
    {
        if(c[i]>='0' && c[i]<='9')
        {
            now*=10;
            now+=c[i] - '0';
        }
        else if(c[i] == '.') 
        {
            st[++top] = now;
            now = 0;
        }
        else if(c[i] == '+')
        {
            st[top-1] = st[top-1] + st[top];
            st[top] = 0;
            top--;
        }
        else if(c[i] == '-')
        {
            st[top-1] = st[top-1] - st[top];
            st[top] = 0;
            top--;
        }
        else if(c[i] == '*')
        {
            st[top-1] = st[top-1] * st[top];
            st[top] = 0;
            top--;
        }
        else if(c[i] == '/')
        {
            st[top-1] = st[top-1] / st[top];
            st[top] = 0;
            top--;
        }
        i++;
    }   
}           
signed main()
{
    scanf("%s",c);
    comp(c);
    printf("%lld\n",st[1]);
    return 0;
}