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。

 1 #include <fstream>
 2 #include <iostream>
 3 #include <vector>
 4 #include <string>
 5 #include <stack>
 6 
 7 using namespace std;
 8 
 9 //#define OJ
10 
11 #ifdef OJ
12 #define fin cin
13 #else
14 ifstream fin("in.txt");
15 #endif
16 
17 struct Node{
18     int cur_address;
19     int val;
20     int next_address;
21 };
22 
23 int main(){
24     string start;
25     int    n, k;
26     fin >> start >> n >> k;
27 
28     vector<Node> nodes(100000);
29 
30     for (int i = 0; i < n; i++){
31         Node tmp;
32         string s1, s2;
33 
34         fin >> s1 >> tmp.val >> s2;
35         tmp.cur_address = atoi(s1.c_str());
36         tmp.next_address = atoi(s2.c_str());
37 
38         nodes[tmp.cur_address] = tmp;
39     }
40 
41     int start_idx = atoi(start.c_str());
42     int cur_address = start_idx;
43     vector<int> path;
44 
45     while (start_idx != -1){
46         path.push_back(start_idx);
47         start_idx = nodes[start_idx].next_address;
48     }
49 
50     vector<int> res;
51     int idx = 0;
52     n = path.size();
53     for (; idx + k - 1 < n; idx += k){
54         stack<int> node_stack;
55         for (int i = 0; i < k; i++){
56             node_stack.push(path[idx + i]);
57         }
58         while (!node_stack.empty()){
59             int tmp = node_stack.top();
60             res.push_back(tmp);
61             node_stack.pop();
62         }
63     }
64     for (; idx < n; idx++){
65         res.push_back(path[idx]);
66     }
67 
68     for (int i = 0; i < n - 1; i++){
69         printf("%05d %d %05d
", nodes[res[i]].cur_address, nodes[res[i]].val, nodes[res[i + 1]].cur_address);
70     }
71     printf("%05d %d %d
", nodes[res[n - 1]].cur_address, nodes[res[n - 1]].val, -1);
72 
73     return 0;
74 }