Digit Counting UVA - 1225
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
HINT
直接暴力破解~
Accepted
#include<stdio.h>
#include<string.h>
int main()
{
int sum;
scanf("%d", &sum);
while (sum--)
{
int num;
int arr[10] = { 0 };
scanf("%d", &num);
for (int i = 1;i <= num;i++)
{
int a, b;
a = i;
while (a)
{
b = a % 10;
a /= 10;
arr[b]++;
}
}
for (int i = 0;i < 9;i++)
printf("%d ", arr[i]);
printf("%d\n", arr[9]);
}
}
Digit Counting UVA - 1225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 ...
- UVa1587.Digit Counting
题目连接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=247&p ...
- 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& ...
- 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 ...
- 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 ...
- 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 ...
- Triangle Counting UVA - 11401(递推)
大白书讲的很好.. #include <iostream> #include <cstring> using namespace std; typedef long long ...
随机推荐
- day1 分布式基础概念
1. 分布式:一个业务分拆多个子业务,部署在不同的服务器上集群:同一个业务,部署在多个服务器上节点:集群中的一个服务器 2.远程调用 分布式系统中调用其它主机 springcloud用http+jso ...
- vue:表格中多选框的处理
效果如下: template中代码如下: <el-table v-loading="listLoading" :data="list" element-l ...
- Elasticsearch 7.x配置用户名密码访问 开启x-pack验证
一.修改elasticsearch 配置文件 1.在配置文件中开启x-pack验证 #进入es安装目录下的config目录 vim elasticsearch.yml # 配置X-Pack http. ...
- Linux内核的TCP协议栈和内核旁路的选择?
[前言]最近在实习公司用到了solarflare的万兆网卡,用到了网卡的openonload技术还有TCPDirect模式代码的编写,其理论基础都是内核旁路.网上关于内核旁路技术的介绍基本就两篇,我结 ...
- Snort + Barbyard2 + Snorby环境搭建
1.环境 ubuntu-14.04.5 daq-2.0.7 Snort-2.9.15.1 Barbyard2 snorby Mysql Docker 2.架构 3.安装步骤 Ubuntu配置 如果是刚 ...
- MySql_176. 第二高的薪水 + limit + distinct + null
MySql_176. 第二高的薪水 LeetCode_MySql_176 题目描述 题解分析 代码实现 # Write your MySQL query statement below select( ...
- TransactionScope 事务
一.TransactionScope是.Net Framework 2.0之后,新增了一个名称空间.它的用途是为数据库访问提供了一个"轻量级"[区别于:SqlTransaction ...
- Java基础语法学习
Java基础语法学习 1. 注释 单行注释: //单行注释 多行注释: /*多行注释 多行注释 多行注释 多行注释 */ 2. 关键字与标识符 关键字: Java所有的组成部分都需要名字.类名.变量名 ...
- WPF 基础 - Binding 对数据的转换和校验
1. Binding 对数据的转换和校验 Binding 中,有检验和转换关卡. 1.1 数据校验 源码: namespace System.Windows.Data { public class B ...
- Hi3559AV100 NNIE开发(4)mobilefacenet.cfg参数配置挖坑解决与SVP_NNIE_Cnn实现分析
前面随笔给出了NNIE开发的基本知识,下面几篇随笔将着重于Mobilefacenet NNIE开发,实现mobilefacenet.wk的chip版本,并在Hi3559AV100上实现mobilefa ...