1092. To Buy or Not to Buy (20)

時間限制 100 ms
內存限制 65536 kB
代碼長度限制 16000 B
判題程序 Standard 作者 CHEN, Yue

Eva would like to make a string of beads with her favorite colors so she went to a small shop to buy some beads. There were many colorful strings of beads. However the owner of the shop would only sell the strings in whole pieces. Hence Eva must check whether a string in the shop contains all the beads she needs. She now comes to you for help: if the answer is "Yes", please tell her the number of extra beads she has to buy; or if the answer is "No", please tell her the number of beads missing from the string.

For the sake of simplicity, let's use the characters in the ranges [0-9], [a-z], and [A-Z] to represent the colors. For example, the 3rd string in Figure 1 is the one that Eva would like to make. Then the 1st string is okay since it contains all the necessary beads with 8 extra ones; yet the 2nd one is not since there is no black bead and one less red bead.

Figure 1

Input Specification:

Each input file contains one test case. Each case gives in two lines the strings of no more than 1000 beads which belong to the shop owner and Eva, respectively.

Output Specification:

For each test case, print your answer in one line. If the answer is "Yes", then also output the number of extra beads Eva has to buy; or if the answer is "No", then also output the number of beads missing from the string. There must be exactly 1 space between the answer and the number.

Sample Input 1: ppRYYGrrYBR2258 YrR8RrY Sample Output 1: Yes 8 Sample Input 2: ppRYYGrrYB225 YrR8RrY Sample Output 1: No 2 解析：本題同樣采用HASH的思想，直接把數組元素值和數組元素位置之間建立索引，而不使用多重循環。

/*************************************************************************> File Name: 1092.c> Author: YueBo> Mail: yuebowhu@163.com> Created Time: Sun 14 May 2017 12:05:03 PM CST************************************************************************/#include<stdio.h> #include <stdio.h>int main() {int extra = 0, miss = 0, flag = 0;int shop_str[10+26+26];char ch;int i;for (i = 0; i < 10+26+26; i++)shop_str[i] = 0;while ((ch = getchar()) != '\n' && ch != ' '){if (ch >= '0' && ch <= '9')shop_str[ch-'0']++;else if (ch >= 'a' && ch <= 'z')shop_str[10+ch-'a']++;elseshop_str[10+26+ch-'A']++;}if (ch == ' '){while ((ch = getchar()) != '\n');}while ((ch = getchar()) != '\n' && ch != ' '){if (ch >= '0' && ch <= '9'){shop_str[ch-'0']--;if (shop_str[ch-'0'] < 0)flag = 1;}else if (ch >= 'a' && ch <= 'z'){shop_str[10+ch-'a']--;if (shop_str[10+ch-'a'] < 0)flag = 1;}else{shop_str[10+26+ch-'A']--;if (shop_str[10+26+ch-'A'] < 0)flag = 1;}}if (flag == 0){for (i = 0; i < 10+26+26; i++)extra += shop_str[i];printf("Yes %d\n", extra);}else{for (i = 0; i < 10+26+26; i++)if (shop_str[i] < 0)miss += shop_str[i];printf("No %d\n", (-1)*miss);}return 0; }

總結

以上是生活随笔為你收集整理的1092. To Buy or Not to Buy (20)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。