1049. Counting Ones (30) The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Inpu…

1049 Counting Ones (30)(30 分) The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12.…

PAT甲级1049. Counting Ones 题意: 任务很简单:给定任何正整数N,你应该计算从1到N的整数的十进制形式的1的总数.例如,给定N为12,在1,10, 11和12. 思路: <编程之美>2.4. 计算每位出现1的次数.所有的加起来就是答案了. 如果该位为0.如12012的百位数. 说明永远取不到121xx的形式.那么这个就相当于12000以下的数所有的可能.所以就是就是这样的形式 n(1)xx ,n为[0,11]所以就是12 * 100 ,即1200种可能. 如果为1.如12…

1049 Counting Ones (30 分) The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Inp…

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 task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For exam…

1049 Counting Ones (30 分)   The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. I…

The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Input Specification: Each inp…

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+=lef…

要统计1到N之间‘1’的个数,如数11包含2个1.所以当N=12时,答案为5. 思想: 找规律,假设ans[N]表示1到N的‘1’的个数,则有a[100]=(a[10]-1)*9+10+a[10]-1+1; 先打表求出1ek的答案: 然后对N由高到低逐位拆分. 有种情况要特别注意: 当N=100001时,高位出现1时要累加到后面第一个非0位数上. #include<iostream> #include<cstring> #include<cstdio> #include…

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<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> #include<string> #include<vector> using namesp…

题目如下: The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Input Specification: Ea…

题目:https://pintia.cn/problem-sets/994805342720868352/problems/994805430595731456 题意: 给定n,问0~n中,1的总个数是多少. 思路: 问的是总个数,所以不需要考虑重复,只用考虑每一位上的贡献就行了. 将数字分成三部分,left(共i位),now和right(共j位) 如果当前now是0, 那么所有前i位是[0,left)的数字都+1个贡献,这些数一共有$left*10^j$个 如果当前now是[2,9],那么所有…

n位数,总共有0~10^n-1共计10^n个数那么所有数出现的总次数变为n*(10^n)个数1出现的次数便是十分之一,所以n位数中,1出现的次数为n*10^(n-1)知道这一个后,接下来就方便求了. 举个例子就方便理解了 3125 从头到尾for一遍 3:那么便有三组1000以内的:0~999,1000~1999,2000~29991000以内的1的个数为300,所以共有3*300=900但是又因为1000~1999中千位上的1也要算进去,有1000个所以0~2999中总共有900+1000=1…

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<bits/stdc++.h> using namespace std; int main(){ ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); int n; cin>>n; ; ,r=,low_bit=,yushu…

题目 The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Input Specification: Each…

The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Input Specification: Each inp…

The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Input Specification: Each inp…

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 task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For exam…

1049. Counting Ones (30) 时间限制 10 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 task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For examp…

1049. Counting Ones (30) The task is simple: given any positive integer N, you are supposed to count the total number of 1's in the decimal form of the integers from 1 to N. For example, given N being 12, there are five 1's in 1, 10, 11, and 12. Inpu…

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1041. Be Unique (20) 题意 给出 N (<=105)个数(数值范围为 [1, 104]),找到其中不重复的第一个数字.比如给出5 31 5 88 67 88 17 , 答案是 31 . 分析 简单模拟题,开一个大数组int a[10001];,以读入的数为下标,记录 count:a[index] ++;.结果输出第一个存储为 1 的下标:if (a[index] == 1). 1042. Shuffling Machine…

1.简单数学 1008 Elevator (20分) 模拟题 #include <cstdio> #include <cstring> #include <string> #include <algorithm> #include <iostream> #include <map> #include <cmath> #define ll long long #define inf 0x3f3f3f using namesp…

题意:题目大意:给出一个数字n,求1~n的所有数字里面出现1的个数 思路:转自(柳婼 の blog)遍历数字的低位到高位,设now为当前位的数字,left为now左边的所有数字构成的数字,right是now右边的所有数字构成的数字.只需一次次累加对于当前位now来说可能出现1的个数,然后把它们累加即可.a表示当前的个位为1,十位为10,百位为100类推. 对于now,有三种情况: 1.now == 0 : 那么 ans += left * a; 2.now == 1 : ans += left…

在上篇,我了解了基数的基本概念,现在进入Linear Counting算法的学习. 理解颇浅,还请大神指点! http://blog.codinglabs.org/articles/algorithms-for-cardinality-estimation-part-ii.html 它的基本处理方法和上篇中用bitmap统计的方法类似,但是最后要用到一个公式: 说明:m为bitmap总位数,u为0的个数,最后的结果为n的一个估计,且为最大似然估计(MLE). 那么问题来了,最大似然估计是什么东东…

来之不易的2017第一发ac http://poj.org/problem?id=2386 Lake Counting Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 31474   Accepted: 15724 Description Due to recent rains, water has pooled in various places in Farmer John's field, which is repr…

ZOJ3944 People Counting ZOJ3939 The Lucky Week 1.PeopleConting 题意:照片上有很多个人,用矩阵里的字符表示.一个人如下: .O. /|\ (.) 占3*3格子,句号“.”为背景.没有两个人完全重合.有的人被挡住了一部分.问照片上有几个人. 题解: 先弄个常量把3*3人形存起来,然后6个部位依次找,比如现在找头,找到一个头,就把这个人删掉(找这个人的各个部位,如果在该部位位置的不是这个人的身体,就不删),删成句号,疯狂找就行了. 代码:…

#include <stdio.h> #include <malloc.h> #define MAX_STACK 10 ; // define the node of stack typedef struct node { int data; node *next; }*Snode; // define the stack typedef struct Stack{ Snode top; Snode bottom; }*NewStack; void creatStack(NewSt…