codeforces 393A Nineteen

发布时间:2014-10-22 18:55:25编辑:www.fx114.net 分享查询网我要评论
本篇文章主要介绍了"codeforces 393A Nineteen",主要涉及到codeforces 393A Nineteen方面的内容,对于codeforces 393A Nineteen感兴趣的同学可以参考一下。

A. Nineteen time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Alice likes word "nineteen" very much. She has a string s and wants the string to contain as many such words as possible. For that reason she can rearrange the letters of the string. For example, if she has string "xiineteenppnnnewtnee", she can get string "xnineteenppnineteenw", containing (the occurrences marked) two such words. More formally, word "nineteen" occurs in the string the number of times you can read it starting from some letter of the string. Of course, you shouldn't skip letters. Help her to find the maximum number of "nineteen"s that she can get in her string. Input The first line contains a non-empty string s, consisting only of lowercase English letters. The length of string s doesn't exceed 100. Output Print a single integer — the maximum number of "nineteen"s that she can get in her string. Sample test(s) input nniinneetteeeenn output 2 input nneteenabcnneteenabcnneteenabcnneteenabcnneteenabcii output 2 input nineteenineteen output 2 地址:http://codeforces.com/problemset/problem/393/A 题意     很简单的题目,就是求用所给的字母能连成几个连续的nineteen(nineteenineteen) 代码 #include <stdio.h> #include <string.h> int main() { char a[200]; while (gets(a) != NULL) { int len = strlen(a); int n, e, i, t; n = e = i = t = 0; for (int j = 0; j < len; j++){ if (a[j] == 'n') n++; else if (a[j] == 'e') e++; else if (a[j] == 'i') i++; else if (a[j] == 't') t++; } int ans; ans = (n - 1) / 2; // 当nineteen个数为ans时,n至少为2*ans+1,所以ans=(n-1)/2 if (ans > e / 3) ans = e / 3; if (ans > i) ans = i; if (ans > t) ans = t; printf("%d\n", ans); } return 0; }


上一篇:swig c python
下一篇:菜鸟入门:Java语言学习六大要点

相关文章

相关评论

本站评论功能暂时取消,后续此功能例行通知。

一、不得利用本站危害国家安全、泄露国家秘密,不得侵犯国家社会集体的和公民的合法权益,不得利用本站制作、复制和传播不法有害信息!

二、互相尊重,对自己的言论和行为负责。

好贷网好贷款