1052. 卖个萌 (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

萌萌哒表情符号通常由“手”、“眼”、“口”三个主要部分组成。简单起见，我们假设一个表情符号是按下列格式输出的：

[左手]([左眼][口][右眼])[右手]

现给出可选用的符号集合，请你按用户的要求输出表情。

输入格式：

输入首先在前三行顺序对应给出手、眼、口的可选符号集。每个符号括在一对方括号[]内。题目保证每个集合都至少有一个符号，并不超过10个符号；每个符号包含1到4个非空字符。

之后一行给出一个正整数K，为用户请求的个数。随后K行，每行给出一个用户的符号选择，顺序为左手、左眼、口、右眼、右手——这里只给出符号在相应集合中的序号（从1开始），数字间以空格分隔。

输出格式：

对每个用户请求，在一行中输出生成的表情。若用户选择的序号不存在，则输出“Are you kidding me? @\/@”。

输入样例：

[╮][╭][o][~\][/~]  [<][>][╯][╰][^][-][=][>][<][@][⊙]
[Д][▽][_][ε][^]  ...
4
1 1 2 2 2
6 8 1 5 5
3 3 4 3 3
2 10 3 9 3

输出样例：

╮(╯▽╰)╭
<(@Д=)/~
o(^ε^)o
Are you kidding me? @\/@

自己写的下面这个代码,感觉什么问题都没有,但是只有一个测例可以通过,代码如下:

/*************************************************************************> File Name: 1052.c> Author: YueBo> Mail: yuebowhu@163.com> Created Time: Sat 17 Jun 2017 11:12:01 PM CST************************************************************************/#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <memory.h>int main()
{char zifu[3][64] = {0};int index[3][64] = {0};int user[1024][5];int sz[3] = {0};int K;int i, j, k;int val;char buffer[128] = {0};int idx[5] = {0,1,2,1,0};int flag;for (i = 0; i < 3; i++){gets(zifu[i]);}for (i = 0; i < 3; i++){j = 0;k = 0;while (zifu[i][j] != '\0'){if (zifu[i][j] == '['){index[i][k] = j;k++;}if (zifu[i][j] == ']'){zifu[i][j] = 0;}j++; }sz[i] = k;}/*for (i = 0; i < 3; i++){for (j = 0; j < sz[i]; j++){printf("%s", &zifu[i][index[i][j]+1]);}printf("\n");}
*/scanf("%d", &K);for (i = 0; i < K; i++){for (j = 0; j < 5; j++){scanf("%d", &user[i][j]);}}for (i = 0; i < K; i++){flag = 0;memset(buffer, 0, 128);for (j = 0; j < 5; j++){val = user[i][j];if (val > sz[idx[j]]){flag = 1;printf("Are you kidding me? @\\/@");break;}strcat(buffer, &zifu[idx[j]][index[idx[j]][val-1]+1]);strcat(buffer, j == 0 ? "(" : "");strcat(buffer, j == 3 ? ")" : "");}printf("%s", flag == 0 ? buffer : "");printf("\n");}return 0;
}

下面这段代码和自己的思想差不多,只不过博主把内容解析出来单独存了起来,原文参见http://blog.csdn.net/chr1991/article/details/51014817

从下面这个代码中得到的启发是,一个任务要进行适当的分解,一步一步的解决,不要一下子把几步的事情一下了了,这样的代码看起来很简洁,但是自己写的时候困难(调的时候困难),别人理解起来困难,分成一步一步的,不要耍快耍酷,怎么清楚明白怎么来.

具体到本题就是:

第1步:把眼,口,手,表情包分别解析出来,存到单独的内存中;

第2步:根据用户请求直接把表情调出来;

#include <stdio.h>  
#define MAX 5  
#define MAX_CH 10  
void read_expression(char array[][MAX], int* count);  
int main(int argc, const char * argv[]) {  char hand[MAX_CH][MAX], eye[MAX_CH][MAX], month[MAX_CH][MAX];  int k, hand1, eye1, month1, eye2, hand2;  int i, hand_count = 0, eye_count = 0, month_count = 0;  read_expression(hand, &hand_count);  read_expression(eye, &eye_count);  read_expression(month, &month_count);  scanf("%d", &k);  for( i = 0; i < k; i++){  scanf("%d %d %d %d %d", &hand1, &eye1, &month1, &eye2, &hand2);  if(hand1 <1 || hand1 > hand_count || hand2 < 1 || hand2 > hand_count){  printf("Are you kidding me? @\\/@\n");  continue;  }  if(eye1 <1 || eye1 > eye_count || eye2 < 1 || eye2 > eye_count){  printf("Are you kidding me? @\\/@\n");  continue;  }  if(month1 < 1 || month1 > month_count){  printf("Are you kidding me? @\\/@\n");  continue;  }  printf("%s(%s%s%s)%s\n", hand[hand1 - 1], eye[eye1 - 1], month[month1 - 1], eye[eye2 - 1], hand[hand2 - 1]);  }  return 0;  
}  
void read_expression(char array[][MAX], int* count){  int i, j, flag;  char temp;  i = 0;  j = 0;  flag = 0;  while((temp = getchar()) != '\n'){  if(temp == '[' && !flag){  j = 0;  flag = 1;  }  else if( temp == ']' && flag){  array[i][j] = '\0';  i++;  flag = 0;  }  else if(flag){  array[i][j++] = temp;  }  }  *count = i;  
}  #include <stdio.h>  
#define MAX 5  
#define MAX_CH 10  
void read_expression(char array[][MAX], int* count);  
int main(int argc, const char * argv[]) {  char hand[MAX_CH][MAX], eye[MAX_CH][MAX], month[MAX_CH][MAX];  int k, hand1, eye1, month1, eye2, hand2;  int i, hand_count = 0, eye_count = 0, month_count = 0;  read_expression(hand, &hand_count);  read_expression(eye, &eye_count);  read_expression(month, &month_count);  scanf("%d", &k);  for( i = 0; i < k; i++){  scanf("%d %d %d %d %d", &hand1, &eye1, &month1, &eye2, &hand2);  if(hand1 <1 || hand1 > hand_count || hand2 < 1 || hand2 > hand_count){  printf("Are you kidding me? @\\/@\n");  continue;  }  if(eye1 <1 || eye1 > eye_count || eye2 < 1 || eye2 > eye_count){  printf("Are you kidding me? @\\/@\n");  continue;  }  if(month1 < 1 || month1 > month_count){  printf("Are you kidding me? @\\/@\n");  continue;  }  printf("%s(%s%s%s)%s\n", hand[hand1 - 1], eye[eye1 - 1], month[month1 - 1], eye[eye2 - 1], hand[hand2 - 1]);  }  return 0;  
}  
void read_expression(char array[][MAX], int* count){  int i, j, flag;  char temp;  i = 0;  j = 0;  flag = 0;  while((temp = getchar()) != '\n'){  if(temp == '[' && !flag){  j = 0;  flag = 1;  }  else if( temp == ']' && flag){  array[i][j] = '\0';  i++;  flag = 0;  }  else if(flag){  array[i][j++] = temp;  }  }  *count = i;  
}

/*************************************************************************> File Name: 1052_1.c> Author: YueBo> Mail: yuebowhu@163.com> Created Time: Tue 20 Jun 2017 08:42:52 PM CST************************************************************************/#include <stdio.h>
#include <stdlib.h>void read_expression(char array[][16], int* count){    char ch;    int i = 0;    int j = 0;    while((ch = getchar()) != '\n'){    if(ch == '['){    i++;j = 0;    }    else if( ch == ']'){    array[i-1][j] = '\0';    j = 0;}    else{    array[i-1][j] = ch;    j++;}    }    *count = i;    
} int GetExpression(char arr[][16])
{char ch = 0;int cnt = 0;int loopi = 0;while ((ch = getchar()) != '\n'){if (ch == '['){cnt++;loopi = 0;continue;}else if (ch == ']'){arr[cnt-1][loopi] = '\0';loopi = 0;}else{arr[cnt-1][loopi] = ch;loopi++;}}return cnt;
}int main()
{char hands[16][16];char eyes[16][16];char mouse[16][16];int cnt;int loopi;//    cnt = GetExpression(hands);read_expression(hands, &cnt);for (loopi = 0; loopi < cnt; loopi++){int i = 0;while (hands[loopi][i] != '\0'){printf("%c", hands[loopi][i]);i++;}}printf("\n");return 0;
}