UVa1587.Digit Counting
| 13764622 | 1225 | Digit Counting | Accepted | C++11 | 0.035 | 2014-06-18 07:44:02 |
1225 - Digit Counting
Time limit: 3.000 seconds
Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequence of consecutive integers starting with 1 to N (1 < N < 10000) . After that, he counts the number of times each digit (0 to 9) appears in the sequence. For example, with N = 13 , the sequence is:
12345678910111213
In this sequence, 0 appears once, 1 appears 6 times, 2 appears 2 times, 3 appears 3 times, and each digit from 4 to 9 appears once. After playing for a while, Trung gets bored again. He now wants to write a program to do this for him. Your task is to help him with writing this program.
Input
The input file consists of several data sets. The first line of the input file contains the number of data sets which is a positive integer and is not bigger than 20. The following lines describe the data sets.
For each test case, there is one single line containing the number N .
Output
For each test case, write sequentially in one line the number of digit 0, 1,...9 separated by a space.
Sample Input
2
3
13
Sample Output
0 1 1 1 0 0 0 0 0 0
1 6 2 2 1 1 1 1 1 1
题目大意:输入一个数N,然后会 123456789……N 这么一个数,算出其中0~9数字出现的个数。
解题思路:题目不难,字符串处理就能轻松解决。主意输出不要有多的空格就ok。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <algorithm>
#include <numeric>
#include <string>
#include <sstream>
using namespace std; int main() {
int l, maxx_int; cin >> l;
string maxx_str;
int tab[];
while(l--) {
cin >> maxx_int;
memset(tab, , sizeof(tab));
for(int i = ; i <= maxx_int; i++) {
stringstream ss;
ss << i; ss >> maxx_str;
for(int j = ; j < maxx_str.size(); j++) {
tab[maxx_str[j] - ''] ++;
}
}
for(int i = ; i < ; i++) {
if(i == ) {
cout << tab[i];
} else { cout << " " << tab[i];
}
}
cout << endl;
//cout << maxx_str << endl;
}
return ;
}
AC代码
UVa1587.Digit Counting的更多相关文章
- UVa 1225 Digit Counting --- 水题
UVa 1225 题目大意:把前n(n<=10000)个整数顺次写在一起,12345678910111213...,数一数0-9各出现多少字 解题思路:用一个cnt数组记录0-9这10个数字出现 ...
- UVA1225 - Digit Counting(紫书习题3.3)
Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequen ...
- Digit Counting UVA - 1225
Trung is bored with his mathematics homeworks. He takes a piece of chalk and starts writing a sequ ...
- 数数字 (Digit Counting,ACM/ICPC Danang 2007,UVa 1225)
思路: 利用java 特性,将数字从1 一直加到n,全部放到String中,然后依次对strring扫描每一位,使其carr[str.charAt(i)-'0']++; 最后输出carr[i],即可. ...
- UVa 1225 Digit Counting
题意:给出n,将前n个整数顺次写在一起,统计各个数字出现的次数. 用的最笨的办法--直接统计-- 后来发现网上的题解有先打表来做的 #include<iostream> #include& ...
- UVa1225 Digit Counting
#include <stdio.h>#include <string.h> int main(){ int T, N, i, j; int a[10]; sc ...
- 数数字(Digit Counting,ACM/ICPC Danang 2007,UVa1225)
#include<stdio.h>#include<stdlib.h>#include<string.h>int main(){ char s[10000]; in ...
- 数数字 (Digit Counting,ACM/ICPC Dannang 2007 ,UVa1225)
题目描述:算法竞赛入门经典习题3-3 #include <stdio.h> #include <string.h> int main(int argc, char *argv[ ...
- UVa 1225 - Digit Counting - ACM/ICPC Danang 2007 解题报告 - C语言
1.题目大意 把前n$(n\le 10000)$个整数顺次写在一起:12345678910111213……计算0~9各出现了多少次. 2.思路 第一想法是打表,然而觉得稍微有点暴力.不过暂时没有想到更 ...
随机推荐
- 【POJ2777】Count Color(线段树)
以下是题目大意: 有水平方向上很多块板子拼成的墙,一开始每一块都被涂成了颜色1,有C和P两个操作,代表的意思是:C X Y Z —— 从X到Y将板子涂成颜色ZP X Y —— 查询X到Y的板子共 ...
- global.asax?app.config?webconfig??
一.Global.asax 1.global.asax是什么? 一个文本文件,至于它包含写什么内容?顾名思义,global 肯定是掌管一个应用程序(application)的全局性的东西,例如应用程序 ...
- Map的内容按字母顺序排序
map有自带的排序功能,但需要重写排序方法,代码如下: package coreJava.com.shindo.corejava.map; import java.util.ArrayList; im ...
- Operation System - Peterson's Solution算法 解决多线程冲突
Person's solution 是用来一种基于软件的解决关键区域问题的算法(critical-section). 它并不是完美的,有可能不对地工作.并且是限制解决两个进程同步的问题. 可是它非常e ...
- 关于时间的操作(JavaScript版)——年月日三级级联(默认依次显示请选择年、请选择月和请选择日)
这篇博客和前一篇博客基本同样,仅仅是显示的默认值不同: <!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN&quo ...
- Java学习笔记——JDBC之与数据库MySQL的连接以及增删改查等操作
必须的准备工作 一.MySQL的安装.可以参考博文: http://blog.csdn.net/jueblog/article/details/9499245 二.下载 jdbc 驱动.可以从在官网上 ...
- FTS下载地址
http://download.microsoft.com/download/5/2/e/52e22b90-2ba7-427b-9ea4-604d3b37a2e7/vs2012_tfs_chs.iso
- 简单JS多级下拉框无刷新
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...
- WCF入门教程系列五
一.概述 WCF在通信过程中有三种模式:请求与答复.单向.双工通信.以下我们一一介绍. 二.请求与答复模式 描述: 客户端发送请求,然后一直等待服务端的响应(异步调用除外),期间处于假死状态,直到服务 ...
- 一个经试用效果非常不错的数据库连接池--JAVA
前言: 虽说现在许多企业级的应用服务器均自己带有数据库连接池功能,就连 Tomcat 也支持了这种功能.然而在许多时候,我们还是要使用数据库连接池,如:访问数据库的 Java 桌面应用程序等.这个数据 ...