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1050. String Subtraction (20)
时间限制10 ms
内存限制65536 kB
代码长度限制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.
排除s2中字符输出即可
#include <cstdio>
#include <cstring>
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;
}
1050. String Subtraction (20)【字符串处理】——PAT (Advanced Level) Practise
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原文地址:http://blog.csdn.net/xianyun2009/article/details/50198525