pat 1049. Counting Ones (30)
看别人的题解懂了一些些 参考《编程之美》P132 页《1 的数目》
#include<iostream>
#include<stdio.h>
using namespace std;
int getone(int n)
{
int ans=0,base=1,right,left,now;
while(n/base)
{
right=n%base;
left=n/(base*10);
now=(n/base)%10;
if(now==0)ans+=left*base;
else
if(now==1)ans+=left*base+right+1;
else
ans+=(left+1)*base;
base*=10;
}
return ans;
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
printf("%d\n",getone(n));
}
return 0;
}
pat 1049. Counting Ones (30)的更多相关文章
- PAT 解题报告 1049. Counting Ones (30)
1049. Counting Ones (30) The task is simple: given any positive integer N, you are supposed to count ...
- pat 甲级 1049. Counting Ones (30)
1049. Counting Ones (30) 时间限制 100 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue The tas ...
- PAT 甲级 1049 Counting Ones (30 分)(找规律,较难,想到了一点但没有深入考虑嫌麻烦)***
1049 Counting Ones (30 分) The task is simple: given any positive integer N, you are supposed to co ...
- PAT 1049 Counting Ones[dp][难]
1049 Counting Ones (30)(30 分) The task is simple: given any positive integer N, you are supposed to ...
- PAT 1049 Counting Ones [难]
1049 Counting Ones (30 分) The task is simple: given any positive integer N, you are supposed to coun ...
- PAT 1004 Counting Leaves (30分)
1004 Counting Leaves (30分) A family hierarchy is usually presented by a pedigree tree. Your job is t ...
- 【PAT甲级】1049 Counting Ones (30 分)(类似数位DP思想的模拟)
题意: 输入一个正整数N(N<=2^30),输出从1到N共有多少个数字包括1. AAAAAccepted code: #define HAVE_STRUCT_TIMESPEC #include& ...
- PAT (Advanced Level) 1049. Counting Ones (30)
数位DP.dp[i][j]表示i位,最高位为j的情况下总共有多少1. #include<iostream> #include<cstring> #include<cmat ...
- PAT Advanced 1049 Counting Ones (30) [数学问题-简单数学问题]
题目 The task is simple: given any positive integer N, you are supposed to count the total number of 1 ...
随机推荐
- Swift - 03 - 整数类型
//: Playground - noun: a place where people can play import UIKit var str = "Hello, playground& ...
- 从头开始学c++,补基础,补踏实
在对c++一知半解的情况下,写c++程序是非常吃力的.对于半路出家写c++的我,写了几个颓废的程序后,再也没有勇气用现有的c++知识去写千疮百孔的程序.非常想写出<整洁的代码>中那样的代码 ...
- 关闭C#主窗体弹出是否关闭对话框
在开发系统时,常常有这样一个问题,就是当关闭主窗体,也即退出系统时,如果想提示是否关闭,以免误操作,可以在主窗体的Main_FormClosing事件中添加一个对话框,代码如下: private vo ...
- underscorejs-find学习
2.5 find 2.5.1 语法: _.find(list, predicate, [context]) 2.5.2 说明: 对list集合的每个成员依次进行匹配(根据predicate迭代函数检测 ...
- WINFORM的DataGridView使用点滴
1.如果有超过一屏的数据,想通过关键字查找到相关记录,然后再定位之,做法如下:遍历所有行,把某单元格的值和关键字对比,找到后清除所有选择行,然后把当前行设为选择,然后把grid的CurrentCell ...
- linux下php上传文件注意
linux下php上传文件注意1.修改上传目录权限linux 修改某目录下所有所有子目录权限chmod -R 777 html修改某目录为任何用户都用写读执行权限chmod a+rwx html2.设 ...
- VueJs一些资料网站链接
https://github.com/liukaijv/laravel-vue-cmshttps://github.com/allan2coder/VUE2-SPA-Tutorialhttps://g ...
- SQLServer 取小时
select datepart(hh,getdate())--orselect datename(hh,getdate())
- cf C. Hamburgers
http://codeforces.com/contest/371/problem/C 二分枚举最大汉堡包数量就可以. #include <cstdio> #include <cst ...
- SCALA当的trait
不是特别懂,但感觉和RUBY当中的MIX-IN功能有几分相似,这又扯到了多重继承及JAVA当中的接口虚拟类了.. package com.hengheng.scala class UseTrait { ...