题目描述:
以上是朋友圈中一奇葩贴:“2月14情人节了,我决定造福大家。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。
输入格式:
输入按照点赞的先后顺序给出不知道多少个点赞的人名,每个人名占一行,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。
输出格式:
根据点赞情况在一行中输出结论:若存在第2个人A和第14个人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 ...”。
输入样例1:
GaoXZh
Magi
Einst
Quark
LaoLao
FatMouse
ZhaShen
fantacy
latesum
SenSen
QuanQuan
whatever
whenever
Potaty
hahaha
.
输出样例1:
Magi and Potaty are inviting you to dinner...
输入样例2:
LaoLao
FatMouse
whoever
.
输出样例2:
FatMouse is the only one for you...
输入样例3:
LaoLao
.
输出样例3:
Momo... No one is for you ...
解题报告:
1:开一个计数器,定义两个字符串,更新就行了。
2:题目规定输入不为空, 字符串初始化为空串。
代码:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main(){
string people2 = "", people14 = "", people;
ll cnt = 0;
while(cin >> people){
if(people == ".")break;
cnt++;
if(cnt == 2)people2 = people;
if(cnt == 14)people14 = people;
}
if(people2 != "" && people14 != "")cout << people2 << " and " << people14 << " are inviting you to dinner...\n";
else if(people2 != "" && people14 == "")cout << people2 << " is the only one for you...\n";
else printf("Momo... No one is for you ...\n");
return 0;
}