第一章字符串处理 4 AcWing 1519. 密码

爱被打了一巴掌 2024-03-31 13:59 31阅读 0赞

## 第一章 字符串处理 4 AcWing 1519. 密码 ##

### 原题链接 ###

[AcWing 1519. 密码][AcWing 1519.]

### 算法标签 ###

模拟 字符串处理

### 思路 ###

依题意模拟

### 代码 ###

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #include<bits/stdc++.h>
    #define int long long
    #define xx first
    #define yy second
    #define ump unordered_map
    #define us unordered_set
    #define pq priority_queue
    #define pb push_back
    #define rep(i, a, b) for(int i=a;i<b;++i)
    #define Rep(i, a, b) for(int i=a;i>=b;--i)
    using namespace std;
    typedef pair<int, int> PII;
    const int N=1005, inf=0x3f3f3f3f3f3f3f3f, mod=1e9+7;
    const double Exp=1e-8;
    //int t, n, m, cnt, ans;
    string na[N], pwd[N];
    char cn[15], cp[15];
    string sn, sp;
    inline int rd(){
       int s=0,w=1;
       char ch=getchar();
       while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
       while(ch>='0'&&ch<='9') s=s*10+ch-'0',ch=getchar();
       return s*w;
    }
    void put(int x) {
        if(x<0) putchar('-'),x=-x;
        if(x>=10) put(x/10);
        putchar(x%10^48);
    }
    string ch(string sp){
        string res="";
        rep(i, 0, sp.size()){
            if(sp[i]=='1'){
                res+="@";
            }else if(sp[i]=='0'){
                res+="%";
            }else if(sp[i]=='l'){
                res+="L";
            }else if(sp[i]=='O'){
                res+="o";
            }else{
                res+=sp[i];
            }
        }
        return res;
    }
    signed main(){
    	ios::sync_with_stdio(false);
    	cin.tie(0);
    	cout.tie(0);
    	int n, m=0;
    	scanf("%lld", &n);
    	rep(i, 0, n){
    	    scanf("%s %s", cn, cp);
    	    sn=cn, sp=cp;
    	    string ssp=ch(sp);
    	    if(ssp!=sp){
    	    
    	        na[m]=sn;
    	        pwd[m]=ssp;
    	        m++;
    	        // 等价于
    	        // na[m]=sn;
    	        // pwd[m++]=ssp;
    	        // 不等价于
    	        // na[m++]=sn;
    	        // pwd[m]=ssp;
    	    }
    	}
    	if(!m){
    	    if(n==1){
    	        puts("There is 1 account and no account is modified");
    	    }else{
    	        printf("There are %lld accounts and no account is modified", n);
    	    }
    	}else{
    	    printf("%lld\n", m);
    	    rep(i, 0, m){
    	        printf("%s %s\n", na[i].c_str(), pwd[i].c_str());
    	    }
    	}
    	return 0;
    }

### 参考文献 ###

[AcWing 1519. 密码（PAT甲级辅导课）y总视频讲解][AcWing 1519. _PAT_y]

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[AcWing 1519.]: https://www.acwing.com/problem/content/1521/
[AcWing 1519. _PAT_y]: https://www.acwing.com/video/983/
[9cb4cb80a18345158f8a69e2f196e5b9.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/31/d6303ec8fd5d41c2b56f93c14dfa64a7.png