PAT 1073

链接：http://www.patest.cn/contests/pat-a-practise/1073

明天就考试了，圆神保佑萌萌哒~

scientific notation，科学记数法，分几种情况进行判断，主要是exponent的部分，如果exponent小于0的话，需要在前面添加0，如果exponent大于0的话，可能在后面添加0.想清楚了不难的。

#include <fstream>
#include <iostream>
#include <vector>
#include <string>
#include <stack>

using namespace std;

//#define OJ

#ifdef OJ
#define fin cin
#else
ifstream fin("in.txt");
#endif

struct Node{
    int cur_address;
    int val;
    int next_address;
};

int main(){
    string start;
    int    n, k;
    fin >> start >> n >> k;

    vector<Node> nodes(100000);

    for (int i = 0; i < n; i++){
        Node tmp;
        string s1, s2;

        fin >> s1 >> tmp.val >> s2;
        tmp.cur_address = atoi(s1.c_str());
        tmp.next_address = atoi(s2.c_str());

        nodes[tmp.cur_address] = tmp;
    }

    int start_idx = atoi(start.c_str());
    int cur_address = start_idx;
    vector<int> path;

    while (start_idx != -1){
        path.push_back(start_idx);
        start_idx = nodes[start_idx].next_address;
    }

    vector<int> res;
    int idx = 0;
    n = path.size();
    for (; idx + k - 1 < n; idx += k){
        stack<int> node_stack;
        for (int i = 0; i < k; i++){
            node_stack.push(path[idx + i]);
        }
        while (!node_stack.empty()){
            int tmp = node_stack.top();
            res.push_back(tmp);
            node_stack.pop();
        }
    }
    for (; idx < n; idx++){
        res.push_back(path[idx]);
    }

    for (int i = 0; i < n - 1; i++){
        printf("%05d %d %05d
", nodes[res[i]].cur_address, nodes[res[i]].val, nodes[res[i + 1]].cur_address);
    }
    printf("%05d %d %d
", nodes[res[n - 1]].cur_address, nodes[res[n - 1]].val, -1);

    return 0;
}