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  • 团体程序设计天梯赛-练习集-L1-035. 情人节

    L1-035. 情人节

    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 ...
    
     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 typedef long long LL;
     4 const int maxn = 100010;
     5 int main() {
     6     string s;
     7     int i = 0;
     8     string a, b;
     9     int flag0 = 0, flag1 = 0;
    10     while (cin >> s) {
    11         if (s == ".")
    12             break;
    13         i += 1;
    14         if (i == 2) {
    15             flag0 = 1;
    16             a = s;
    17         }
    18         if (i == 14) {
    19             b = s;
    20             flag1 = 1;
    21         }
    22     }
    23     if (flag0 == 0 && flag1 == 0) {
    24         printf ("Momo... No one is for you ...
    ");
    25     } else if (flag0 == 1 && flag1 == 0) {
    26         cout << a;
    27         printf (" is the only one for you...
    ");
    28     } else {
    29         cout << a << " and " << b;
    30         printf (" are inviting you to dinner...
    ");
    31     }
    32     return 0;
    33 }
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  • 原文地址:https://www.cnblogs.com/zhien-aa/p/8612234.html
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