标签:col lan 附加 void fir iostream 号码 for set
给定大量手机用户通话记录,找出其中通话次数最多的聊天狂人。
输入首先给出正整数N(≤10?^5??),为通话记录条数。随后N行,每行给出一条通话记录。简单起见,这里只列出拨出方和接收方的11位数字构成的手机号码,其中以空格分隔。
在一行中给出聊天狂人的手机号码及其通话次数,其间以空格分隔。如果这样的人不唯一,则输出狂人中最小的号码及其通话次数,并且附加给出并列狂人的人数。
4
13005711862 13588625832
13505711862 13088625832
13588625832 18087925832
15005713862 13588625832
13588625832 3
/* 6 13005711862 13588625832 13505711862 13088625832 13588625832 18087925832 15005713862 13588625832 13005711862 15005713862 13005711862 15005713862 */ #include<iostream> #include<string> #include<map> using namespace std; const int maxn = 100010; map<string, int> mp; void Input(string str) { if (mp.find(str) == mp.end()) { mp[str] = 1; } else { mp[str]++; } } int main() { int n; cin >> n; string s1,s2; for (int i = 0; i < n; i++) { cin >> s1 >> s2; Input(s1); Input(s2); } map<string, int>::iterator it = mp.begin(); int most_time = -1; int most_num = 0; string most_ID; for (it = mp.begin(); it != mp.end(); it++) { //cout << "it->second = " << it->second << endl; if (it->second > most_time) { most_time = it->second; most_ID = it->first; most_num = 1; } else if(it->second == most_time ) { most_num++; if (most_ID > it->first) { most_ID = it->first; } } } if (1 == most_num) { cout << most_ID << most_time; } else { cout << most_ID << " "<< most_time << " "<< most_num; } return 0; }
标签:col lan 附加 void fir iostream 号码 for set
原文地址:https://www.cnblogs.com/wanghao-boke/p/11960779.html