立志用最少的代碼做最高效的表達(dá)

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PAT甲級(jí)最優(yōu)題解——>傳送門

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To prepare for PAT, the judge sometimes has to generate random passwords for the users. The problem is that there are always some confusing passwords since it is hard to distinguish 1 (one) from l (L in lowercase), or 0 (zero) from O (o in uppercase). One solution is to replace 1 (one) by @, 0 (zero) by %, l by L, and O by o. Now it is your job to write a program to check the accounts generated by the judge, and to help the juge modify the confusing passwords.

Input Specification:
Each input file contains one test case. Each case contains a positive integer N (≤1000), followed by N lines of accounts. Each account consists of a user name and a password, both are strings of no more than 10 characters with no space.

Output Specification:
For each test case, first print the number M of accounts that have been modified, then print in the following M lines the modified accounts info, that is, the user names and the corresponding modified passwords. The accounts must be printed in the same order as they are read in. If no account is modified, print in one line There are N accounts and no account is modified where N is the total number of accounts. However, if N is one, you must print There is 1 account and no account is modified instead.

Sample Input 1:
3
Team000002 Rlsp0dfa
Team000003 perfectpwd
Team000001 R1spOdfa
Sample Output 1:
2
Team000002 RLsp%dfa
Team000001 R@spodfa

Sample Input 2:
1
team110 abcdefg332
Sample Output 2:
There is 1 account and no account is modified

Sample Input 3:
2
team110 abcdefg222
team220 abcdefg333
Sample Output 3:
There are 2 accounts and no account is modified

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水題，注意點(diǎn)（測(cè)試點(diǎn)2）：

若所有密碼都沒有被替換的字符：

如果N為1， 那么輸出There is 1 account and no account is modified

反之，輸出There are N accounts and no account is modified

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#include<bits/stdc++.h> using namespace std; struct teamer{string id, password;bool change = false; }ter[1010]; int main() {int n; cin >> n;for(int i = 0; i < n; i++) cin >> ter[i].id >> ter[i].password;int num_pass = 0; //被修改字符總次數(shù)、被修改的密碼數(shù) for(int i = 0; i < n; i++) {for(auto& j : ter[i].password) {/**/ if(j == '1') { j = '@'; ter[i].change = true; } //沒錯(cuò)我就是強(qiáng)迫癥 else if(j == '0') { j = '%'; ter[i].change = true; }else if(j == 'l') { j = 'L'; ter[i].change = true; }else if(j == 'O') { j = 'o'; ter[i].change = true; }}if(ter[i].change) num_pass++; } //輸出 if(num_pass == 0) cout << "There " << (n==1?"is ":"are ") << n << " account" << (n==1?"":"s") << " and no account is modified\n";else {cout << num_pass << '\n';for(int i = 0; i < n; i++) if(ter[i].change) cout << ter[i].id << ' ' << ter[i].password << '\n';} return 0; }

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耗時(shí)：

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??????????——甲級(jí)考試倒計(jì)時(shí)12days

超強(qiáng)干貨來(lái)襲 云風(fēng)專訪：近40年碼齡，通宵達(dá)旦的技術(shù)人生

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