A registration card number of PAT consists of 4 parts:
the 1st letter represents the test level, namely, T for the top level, A for advance and B for basic;the 2nd - 4th digits are the test site number, ranged from 101 to 999;the 5th - 10th digits give the test date, in the form of yymmdd;finally the 11th - 13th digits are the testee’s number, ranged from 000 to 999.Now given a set of registration card numbers and the scores of the card owners, you are supposed to output the various statistics according to the given queries. Input Specification: Each input file contains one test case. For each case, the first line gives two positive integers N (≤104) and M (≤100), the numbers of cards and the queries, respectively. Then N lines follow, each gives a card number and the owner’s score (integer in [0,100]), separated by a space. After the info of testees, there are M lines, each gives a query in the format Type Term, where Type being 1 means to output all the testees on a given level, in non-increasing order of their scores. The corresponding Term will be the letter which specifies the level; Type being 2 means to output the total number of testees together with their total scores in a given site. The corresponding Term will then be the site number; Type being 3 means to output the total number of testees of every site for a given test date. The corresponding Term will then be the date, given in the same format as in the registration card. Output Specification: For each query, first print in a line Case #: input, where # is the index of the query case, starting from 1; and input is a copy of the corresponding input query. Then output as requested: for a type 1 query, the output format is the same as in input, that is, CardNumber Score. If there is a tie of the scores, output in increasing alphabetical order of their card numbers (uniqueness of the card numbers is guaranteed); for a type 2 query, output in the format Nt Ns where Nt is the total number of testees and Ns is their total score; for a type 3 query, output in the format Site Nt where Site is the site number and Nt is the total number of testees at Site. The output must be in non-increasing order of Nt’s, or in increasing order of site numbers if there is a tie of Nt. If the result of a query is empty, simply print NA.
Sample Input: 8 4 B123180908127 99 B102180908003 86 A112180318002 98 T107150310127 62 A107180908108 100 T123180908010 78 B112160918035 88 A107180908021 98 1 A 2 107 3 180908 2 999 Sample Output: Case 1: 1 A A107180908108 100 A107180908021 98 A112180318002 98 Case 2: 2 107 3 260 Case 3: 3 180908 107 2 123 2 102 1 Case 4: 2 999 NA
思路： 研读柳神的代码，开阔眼界，可以将具有相同结构的代码放在一起，减少了代码量，也减少了出错量
注意：
unordered_map只有在C++11才开始支持直接写结构体{}，也是C++11的特性，auto也是代码：(C++)
#include<iostream> #include<vector> #include<map> #include<unordered_map> //c++ 2011才能用 #include<algorithm> using namespace std; struct node { string t; int value; }; bool cmp(const node &a, const node &b) { return a.value != b.value ? a.value > b.value : a.t < b.t; } int main() { int n,k,num; string s; cin>>n>>k; vector<node> v(n); //先存入各考生信息 for(int i=0; i<n; i++) cin>> v[i].t >> v[i].value; for(int i=1; i<=k; i++) { cin>>num>>s; printf("Case %d: %d %s\n", i, num, s.c_str()); vector<node> ans; //对于本次查找要用到的数据 int cnt = 0, sum = 0; if(num == 1) { for(int j=0; j<n; j++) if(v[j].t[0] == s[0]) ans.push_back(v[j]); } else if(num == 2) { for(int j=0; j<n; j++) { if(v[j].t.substr(1,3) == s) { cnt++; sum += v[j].value; } } if(cnt != 0) printf("%d %d\n", cnt, sum); } else if(num == 3) { // map<string,int> m; unordered_map<string,int> m; for(int j=0; j<n; j++) { if(v[j].t.substr(4,6) == s) m[v[j].t.substr(1,3)]++; } // for(map<string,int>::iterator it=m.begin(); it!=m.end(); it++) for(unordered_map<string,int>::iterator it=m.begin(); it!=m.end(); it++) { node emp; emp.t = it->first; emp.value = it->second; ans.push_back(emp); } } sort(ans.begin(),ans.end(),cmp); //把1和3的排序整合在一起 for(int j=0; j<ans.size(); j++) //输出情况为1,3的，否则没有输出 { printf("%s %d\n", ans[j].t.c_str(), ans[j].value); } if(((num == 1 || num == 3) && ans.size() == 0) || (num == 2 && cnt == 0)) printf("NA\n"); } return 0; }参考： https://blog.csdn.net/liuchuo/article/details/84973049
