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 ...
代码:
#include<stdio.h> #include<string.h> int main() { int i,j,n,m,k,t; char name[15],nameA[15],nameB[15]; k=0; while(scanf("%s",name)&&strcmp(name,".")!=0) { k++; if(k==2) { strcpy(nameA,name); } if(k==14) { strcpy(nameB,name); } } if(k>=14) { printf("%s and %s are inviting you to dinner...",nameA,nameB); } else if(k>=2) { printf("%s is the only one for you...",nameA); } else { printf("Momo... No one is for you ..."); } return 0; }