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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> 1050. String Subtraction (20)[字符串處理]——PAT (Advanced Level) Practise

1050. String Subtraction (20)[字符串處理]——PAT (Advanced Level) Practise

編輯:C++入門知識

1050. String Subtraction (20)[字符串處理]——PAT (Advanced Level) Practise


題目信息

1050. String Subtraction (20)

時間限制10 ms
內存限制65536 kB
代碼長度限制16000 B

Given two strings S1 and S2, S = S1 - S2 is defined to be the remaining string after taking all the characters in S2 from S1. Your task is simply to calculate S1 - S2 for any given strings. However, it might not be that simple to do it fast.

Input Specification:

Each input file contains one test case. Each case consists of two lines which gives S1 and S2, respectively. The string lengths of both strings are no more than 10^4. It is guaranteed that all the characters are visible ASCII codes and white space, and a new line character signals the end of a string.

Output Specification:

For each test case, print S1 - S2 in one line.

Sample Input:
They are students.
aeiou
Sample Output:
Thy r stdnts.

解題思路

排除s2中字符輸出即可

AC代碼

#include 
#include 
char s[10005], aim[10005];

int main()
{
    gets(s);
    gets(aim);
    int len = strlen(s);
    for (int i = 0; i < len; ++i){
        if (NULL == strchr(aim, s[i])){
            putchar(s[i]);
        }
    }
    return 0;
}

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