标签：typedef amp please getch ase coder cte home tor
Heshen was an official of the Qing dynasty. He made a fortune which could be comparable to a whole country‘s wealth by corruption. So he was known as the most corrupt official in Chinese history. But Emperor Qianlong liked, or even loved him so much that was not punished during Qianlong‘s regime even everybody knew he was totally corrupted.
After Qianlong quit his job, his son Jiaqing took the throne. The new Emperor hated Heshen very much, so he threw Heshen into jail and awarded him death immediately after Qianlong died. Of course Jiaqing would not forget to raid Heshen‘s house for money --- that was the story of "Heshen fell, Jiaqing full."
Jiaqing‘s man got a notebook from Heshen‘s home which was obviously an account book.But the text of the book was written in English! Jiaqing thought all numbers in that account book should stand for money. Please find out all numbers for Jiaqing.
The text of the account book consists only lower case letters, spaces, and digits
(‘0‘ - ‘9‘). One or several consecutive digits form a number. But Jiaqing only cared about the ACTUAL numbers among all numbers. Only if a number DOESN‘T satisfy any of the conditions below, it is an ACTUAL number:
1) The character to the left of the number is a lower case letter, for example: a123
2) The character to the right of the number is a lower case letter, for example: 123b
3) The number has one or more extra leading zeros, such as 01 , 0012….
Please note that if the last character of a line is a digit, and the first character of the next line is also a digit, those two digits are considered consecutive.
Input
There are no more than 200 lines. The length of each line is no more than 1000 characters.
And it is guaranteed that every number‘s length is no more than 18.
There may be spaces at the end of a line, and those spaces matter.
No blank lines in the input. A line consisting of only spaces is not a blank line.
Output
Print all ACTUAL numbers in a single line in the original order.
Then, count the number of ACTUAL numbers of each line, and print them. A number X only belongs to the line which contains the first digit of X.
Sample Input
a19 01 17b 12 bdc 13 23 14 344 bc
Sample Output
12 1323 14 344 0 2 2
题意：
给你多行只含有数字，空格，小写字母的字符串。
让你从中输出她们想要的数字，并记录每一行有多少个他们想要的数字。
他们想要的数字有以下限制：
第一个字符不能是字母。
最后一个字符不能是字母。
不能有前导0.
题目有以下坑点（巨坑，毒瘤）：
1，
123edwe124 是一个有效的数字 它是 123124 （心情：出题人你认真的？）
2：最后一行的尾部没有回车 ，处理的时候要额外注意
3：如果一行的最后一个字符是数字，下一行的第一个字符是数字，那么下一行的头部数字是连着上一行的，（奇葩的处理）。
4. 只有一个0， 不是前导0，而是一个有效数字0.
解决坑点就按照题意模拟实现就好了，细节见代码：
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <queue> #include <stack> #include <map> #include <set> #include <vector> #include <iomanip> #define ALL(x) (x).begin(), (x).end() #define rt return #define dll(x) scanf("%I64d",&x) #define xll(x) printf("%I64d\n",x) #define sz(a) int(a.size()) #define all(a) a.begin(), a.end() #define rep(i,x,n) for(int i=x;i<n;i++) #define repd(i,x,n) for(int i=x;i<=n;i++) #define pii pair<int,int> #define pll pair<long long ,long long> #define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0) #define MS0(X) memset((X), 0, sizeof((X))) #define MSC0(X) memset((X), ‘\0‘, sizeof((X))) #define pb push_back #define mp make_pair #define fi first #define se second #define eps 1e-6 #define gg(x) getInt(&x) #define db(x) cout<<"== [ "<<x<<" ] =="<<endl; using namespace std; typedef long long ll; ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} ll powmod(ll a, ll b, ll MOD) {ll ans = 1; while (b) {if (b % 2)ans = ans * a % MOD; a = a * a % MOD; b /= 2;} return ans;} inline void getInt(int* p); const int maxn = 100010; const int inf = 0x3f3f3f3f; /*** TEMPLATE CODE * * STARTS HERE ***/ char s[400010]; int ans[maxn]; bool check(string str) { if (str == "") { return 0; } int len = str.length(); if (str[0] >= ‘a‘ && str[0] <= ‘z‘) { return 0; } else if (str[len - 1] >= ‘a‘ && str[len - 1] <= ‘z‘) { return 0; } else { if (len == 1) { return 1; } else { if (str[0] == ‘0‘) { return 0; } else { return 1; } } } } int main() { // freopen("D:\\common_text\\code_stream\\in.txt","r",stdin); //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout); int n = 0; char x; while (~scanf("%c", &x)) { s[n++] = x; } string temp = ""; int cnt = 1; std::vector<string> v; int huiche = 0; rep(i, 0, n) { if (s[i] == ‘ ‘) { if (check(temp)) { // db(temp); int num = temp.length(); string str = ""; rep(j, 0, num) { if (temp[j] >= ‘0‘ && temp[j] <= ‘9‘) { str += temp[j]; } } v.push_back(str); ans[cnt]++; // cout<<cnt<<" "<<str<<" "<<huiche<<endl; } cnt += huiche; huiche = 0; temp = ""; continue; } else if (s[i] == ‘\n‘) { huiche++; int num = temp.length(); if (s[i - 1] >= ‘0‘ && s[i - 1] <= ‘9‘) { if (s[i + 1] >= ‘0‘ && s[i + 1] <= ‘9‘) { int j = i + 1; while (s[j] != ‘ ‘ && s[j] != ‘\n‘) { temp += s[j++]; } i = j - 1; } else { if (check(temp)) { // db(temp); int num = temp.length(); string str = ""; rep(j, 0, num) { if (temp[j] >= ‘0‘ && temp[j] <= ‘9‘) { str += temp[j]; } } v.push_back(str); ans[cnt]++; // cout<<cnt<<" "<<str<<" "<<huiche<<endl; } cnt += huiche; huiche = 0; temp = ""; continue; } } else { if (check(temp)) { // db(temp); int num = temp.length(); string str = ""; rep(j, 0, num) { if (temp[j] >= ‘0‘ && temp[j] <= ‘9‘) { str += temp[j]; } } v.push_back(str); ans[cnt]++; // cout<<cnt<<" "<<str<<" "<<huiche<<endl; } cnt += huiche; huiche = 0; temp = ""; continue; } } else { temp += s[i]; } } if (check(temp)) { // db(temp); int num = temp.length(); string str = ""; rep(j, 0, num) { if (temp[j] >= ‘0‘ && temp[j] <= ‘9‘) { str += temp[j]; } } v.push_back(str); ans[cnt]++; // cout<<cnt<<" "<<str<<" "<<huiche<<endl; } for (int i = 0; i < sz(v); ++i) { cout << v[i]; if (i != sz(v) - 1) { cout << " "; } } cout << endl; repd(i, 1, cnt) { // cout<<i<<" "<<ans[i]<<endl; cout << ans[i] << endl; } return 0; } inline void getInt(int* p) { char ch; do { ch = getchar(); } while (ch == ‘ ‘ || ch == ‘\n‘); if (ch == ‘-‘) { *p = -(getchar() - ‘0‘); while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 - ch + ‘0‘; } } else { *p = ch - ‘0‘; while ((ch = getchar()) >= ‘0‘ && ch <= ‘9‘) { *p = *p * 10 + ch - ‘0‘; } } }
Heshen's Account Book HihoCoder - 1871 2018北京区域赛B题（字符串处理）
标签：typedef amp please getch ase coder cte home tor
原文地址：https://www.cnblogs.com/qieqiemin/p/10806772.html
