UVA 10361
2015-07-20 20:27
387 查看
A Schuttelreim seems to be a typical German invention. The funny thing about this strange type of poetry is that if somebody gives you the first line and the beginning of the second one, you can complete the poem yourself. Well, even a computer can do that,
and your task is to write a program which completes them automatically. This will help Lara concentrate on the “action” part of Tomb Raider and not on the “intellectual” part.
s1<s2>s3<s4>s5
where the si are possibly empty, strings of lowercase characters or blanks. The second line will be a string of lowercase characters or blanks ending with three dots “...”. Lines will we at most 100 characters long.
For each pair of Schuttelreim lines l1 and l2 you are to output two lines c1 and c2 in the following way: c1 is the same as l1 only that the bracket marks “<” and “>” are removed. Line c2
ein kind haelt seinen< schn>abel <n>ur
wenn es haengt an der ...
weil wir zu spaet zur<> oma <k>amen
verpassten wir das ...
<d>u <b>ist
...
wenn es haengt an der nabel schnur
weil wir zu spaet zur oma kamen
verpassten wir das koma amen
du bist
bu dist
代码#include<stdio.h>
#include<string.h>
#define N 100
int main (){
int i,j,Case,len;
char a
,b
,s
,s1
,s2
,s3
,s4
,s5
;
while(scanf("%d\n",&Case) != EOF){
while(Case--){
gets(a);
gets(b);
len = strlen(a);
for(j = 0,i = 0;a[i] != '<';i++,j++){
s1[j] = a[i];
}
s1[j] = '\0';
for(i = i+1,j = 0;a[i] != '>';i++,j++){
s2[j] = a[i];
}
s2[j] = '\0';
for(i = i+1,j = 0;a[i] != '<';i++,j++){
s3[j] = a[i];
}
s3[j] = '\0';
for(i = i+1,j = 0;a[i] != '>';i++,j++){
s4[j] = a[i];
}
s4[j] = '\0';
for(i = i+1,j = 0;i < len;i++,j++){
s5[j] = a[i];
}
s5[j] = '\0';
for(i = 0,j = 0;b[i] != '.';i++,j++){
s[j] = b[i];
}
s[j] = '\0';
printf("%s%s%s%s%s\n",s1,s2,s3,s4,s5);
printf("%s%s%s%s%s\n",s,s4,s3,s2,s5);
4000
}
}
return 0;
and your task is to write a program which completes them automatically. This will help Lara concentrate on the “action” part of Tomb Raider and not on the “intellectual” part.
Input
The input will begin with a line containing a single number n. After this line follow n pairs of lines containing Schuttelreims. The first line of each pair will be of the forms1<s2>s3<s4>s5
where the si are possibly empty, strings of lowercase characters or blanks. The second line will be a string of lowercase characters or blanks ending with three dots “...”. Lines will we at most 100 characters long.
Output
For each pair of Schuttelreim lines l1 and l2 you are to output two lines c1 and c2 in the following way: c1 is the same as l1 only that the bracket marks “<” and “>” are removed. Line c2
is the same as l2 , except that instead of the three dots the string s4s3s2s5 should appear.
Sample Input
3ein kind haelt seinen< schn>abel <n>ur
wenn es haengt an der ...
weil wir zu spaet zur<> oma <k>amen
verpassten wir das ...
<d>u <b>ist
...
Sample Output
ein kind haelt seinen schnabel nurwenn es haengt an der nabel schnur
weil wir zu spaet zur oma kamen
verpassten wir das koma amen
du bist
bu dist
代码#include<stdio.h>
#include<string.h>
#define N 100
int main (){
int i,j,Case,len;
char a
,b
,s
,s1
,s2
,s3
,s4
,s5
;
while(scanf("%d\n",&Case) != EOF){
while(Case--){
gets(a);
gets(b);
len = strlen(a);
for(j = 0,i = 0;a[i] != '<';i++,j++){
s1[j] = a[i];
}
s1[j] = '\0';
for(i = i+1,j = 0;a[i] != '>';i++,j++){
s2[j] = a[i];
}
s2[j] = '\0';
for(i = i+1,j = 0;a[i] != '<';i++,j++){
s3[j] = a[i];
}
s3[j] = '\0';
for(i = i+1,j = 0;a[i] != '>';i++,j++){
s4[j] = a[i];
}
s4[j] = '\0';
for(i = i+1,j = 0;i < len;i++,j++){
s5[j] = a[i];
}
s5[j] = '\0';
for(i = 0,j = 0;b[i] != '.';i++,j++){
s[j] = b[i];
}
s[j] = '\0';
printf("%s%s%s%s%s\n",s1,s2,s3,s4,s5);
printf("%s%s%s%s%s\n",s,s4,s3,s2,s5);
4000
}
}
return 0;
相关文章推荐
- PAT (Advanced Level) 1014. Waiting in Line (30) 银行排队
- 谈谈八皇后的问题
- poj 1258 Agri-Net(Prim)(基础)
- 画风、设计、ARPG:如何做好下一款三消游戏?
- 剑指off-找到最小k个数字
- 东华软件面试题
- OC中字符串用法总结
- 在CentOS 7中Samba服务安装和配置
- 每天一个小知识点8(jQuer总结二)
- 常见的查找算法
- Codeforces Round #290 (Div. 2)D. Fox And Jumping
- 职业性格测验量表
- tornado autoreload 模式
- igrimace最新安装源 apt.so/igrimace3
- 一个分布式服务器集群架构方案
- Test for Job (poj 3249 记忆化搜索)
- hdu 1178 免费馅饼
- NOJ_1011 大数加法
- expdp数据库时报ora-39002错误问题的处理01
- fmdb