以上是朋友圈中一奇葩贴:“2月14情人节了,我决定造福你们。第2个赞和第14个赞的,我介绍你俩认识…………咱三吃饭…你俩请…”。现给出此贴下点赞的朋友名单,请你找出那两位要请客的倒霉蛋。ios
输入按照点赞的前后顺序给出不知道多少个点赞的人名,每一个人名占一行,为不超过10个英文字母的非空单词,以回车结束。一个英文句点.
标志输入的结束,这个符号不算在点赞名单里。spa
根据点赞状况在一行中输出结论:若存在第2我的A和第14我的B,则输出“A and B are inviting you to dinner...”;若只有A没有B,则输出“A is the only one for you...”;若连A都没有,则输出“Momo... No one is for you ...”。code
GaoXZh Magi Einst Quark LaoLao FatMouse ZhaShen fantacy latesum SenSen QuanQuan whatever whenever Potaty hahaha .
Magi and Potaty are inviting you to dinner...
LaoLao FatMouse whoever .
FatMouse is the only one for you...
LaoLao .
Momo... No one is for you ...
#include<iostream> #include<string> using namespace std; int main() { int cnt = 0; string str, A, B; while (cin >> str) { if (str == ".") break; else { cnt++; if (cnt == 2) A = str; if (cnt == 14) B = str; } } if (cnt >= 14) cout << A << " and " << B << " are inviting you to dinner..."; if (cnt >= 2 && cnt < 14) cout << A << " is the only one for you..."; if (cnt < 2) cout << "Momo... No one is for you ..."; return 0; }