ZOJ - 3930 Dice Notation 【模拟】

题目链接

http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3930

题意
给出一串字符串

如果是 ‘+’ ‘-’ ‘*’ ‘/’ 那么需要在前后分别加一个空格

如果遇到 纯数字 直接输出

如果遇到 adx 这样的

要化成 ([dx] + [dx] + [dx]) (a 个 dx)

但是要注意 x 和上面的纯数字 都可能是大数 然后 a 可以用long long 保存

还有 1dx 或者 dx 都是 输出 [dx] 两端没有括号

AC代码

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair  pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = acos(-1);
const double E = exp(1);
const double eps = 1e-30;

const int INF = 0x3f3f3f3f;
const int maxn = 5e4 + 5;
const int MOD = 1e9 + 7;

ll tran(string s)
{
    ll ans = 0;
    int len = s.size();
    for (int i = 0; i < len; i++)
        ans = ans * 10 + s[i] - '0';
    return ans;
}

int main()
{
    int T;
    scanf("%d ", &T);
    while (T--)
    {
        string s;
        getline(cin, s);
        int len = s.size();
        string temp = "";
        for (int i = 0; i < len; i++)
        {
            if (s[i] == ' ')
                continue;
            else if (s[i] == '(' || s[i] == ')')
                printf("%c", s[i]);
            else if (s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/')
                printf(" %c ", s[i]);
            else if (isdigit(s[i]))
            {
                temp.clear();
                temp = temp + s[i];
                i++;
                while (i < len && isdigit(s[i]))
                    temp = temp + s[i++];
                if (s[i] == 'd')
                {
                    ll num = tran(temp);
                    if (num != 1)
                        printf("(");
                    temp.clear();
                    temp = temp + "[d";
                    i++;
                    while (i < len && isdigit(s[i]))
                        temp = temp + s[i++];
                    i--;
                    temp = temp + ']';
                    int len = temp.size();
                    for (ll i = 0; i < num; i++)
                    {
                        cout << temp;
                        if (i <= num - 2)
                            printf(" + ");
                    }
                    if (num != 1)
                        printf(")");
                }
                else
                {
                    i--;
                    cout << temp;
                }

            }
            else if (s[i] == 'd')
            {
                printf("[d");
                i++;
                while (isdigit(s[i]) && i < len)
                {
                    printf("%c", s[i++]);
                }
                i--;
                printf("]");
            }
        }
        printf(" = [Result]\n");
    }
}

转载于:https://www.cnblogs.com/Dup4/p/9433141.html