PAT (Advanced Level) 1050. String Subtraction (20)

简单题。

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<map>
#include<queue>
#include<string>
#include<vector>
using namespace std;

const int maxn=100000+10;
char t[maxn];
char s[maxn];
int flag[maxn];

int main()
{
    gets(s);
    gets(t);
    memset(flag,0,sizeof flag);
    for(int i=0;t[i];i++) flag[t[i]]=1;
    for(int i=0;s[i];i++)
    {
        if(flag[s[i]]==1) continue;
        printf("%c",s[i]);
    }
    printf("\n");
    return 0;
}
时间： 2024-08-29 18:12:19

PAT (Advanced Level) 1050. String Subtraction (20)的相关文章

【PAT甲级】1050 String Subtraction (20 分)

题意: 输入两个串,长度小于10000,输出第一个串去掉第二个串含有的字符的余串. trick: ascii码为0的是NULL,减去'0','a','A',均会导致可能减成负数. 代码: #define HAVE_STRUCT_TIMESPEC#include<bits/stdc++.h>using namespace std;char s1[10007],s2[10007];int vis[507];int main(){ ios::sync_with_stdio(false); cin.t

1050. String Subtraction (20)【字符串处理】——PAT (Advanced Level) Practise

题目信息 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 giv

PAT Advanced 1050 String Subtraction (20 分)

Given two strings S?1?? and S?2??, S=S?1??−S?2?? is defined to be the remaining string after taking all the characters in S?2?? from S?1??. Your task is simply to calculate S?1??−S?2?? for any given strings. However, it might not be that simple to do

PAT Advanced 1050 String Subtraction (20) [Hash散列]

题目 Given two strings S1 and S2, S = S1 – S2 is defined to be the remaining string afer 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

PAT：1050. String Subtraction (20) AC

#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; bool HARSH[260]; //实际上申请来之后初试都是false int main() { fill(HARSH,HARSH+260,true); char str1[10066]; char str2[10066]; gets(str1); gets(str2); int len1=strlen(str1)

PAT甲题题解-1050. String Subtraction (20)-水题

#include <iostream> #include <cstdio> #include <string.h> #include <algorithm> using namespace std; /* 水题,注意字符范围是整个ASCII编码即可. */ const int maxn=130; int vis[maxn]; char s1[10000+5]; char s2[10000+5]; int main() { gets(s1); //getcha

1050. String Subtraction (20)

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 Sp

PAT (Advanced Level) 1081. Rational Sum (20)

简单模拟题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; struct FenShu { long l

PAT (Advanced Level) 1073. Scientific Notation (20)

简单模拟题. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<map> #include<stack> #include<queue> #include<string> #include<algorithm> using namespace std; char s[100000]; int p;