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
.结尾无空行
输出样例1:
Magi and Potaty are inviting you to dinner...
结尾无空行
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 ...
answer
#include <stdio.h>
#include <string.h>
int main() {
char A[11] = {'\0'};
char B[11] = {'\0'};
char T[11] = {'\0'};
int cnt = 0;
while(1) {
gets(T);
if(T[0] == '.')
break;
cnt++;
if(cnt == 2)
strcpy(A,T);
if(cnt == 14)
strcpy(B,T);
}
if(cnt >= 14)
printf("%s and %s are inviting you to dinner...",A,B);
else if(cnt >= 2 && cnt < 14)
printf("%s is the only one for you...",A);
else
printf("Momo... No one is for you ...");
return 0;
}