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) 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…

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…

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…

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 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…

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.…

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 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…

1004. Counting Leaves (30)   A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100,…

1004. Counting Leaves (30) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100, th…

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…

1004 Counting Leaves (30)(30 分) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 10…

1004. Counting Leaves (30) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one…

1004 Counting Leaves (30分) A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing 0…

1004 Counting Leaves (30分)   A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Specification: Each input file contains one test case. Each case starts with a line containing…

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…

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…

题目链接:https://www.patest.cn/contests/pat-a-practise/1004 大意:输出按层次输出每层无孩子结点的个数 思路:vector存储结点,dfs遍历 #include<iostream> #include<cstdio> #include<string> #include<vector> #include<algorithm> using namespace std; ; int n,m,k,x,f[m…

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…

利用广度优先搜索,找出每层的叶子节点的个数. #include <iostream> #include <vector> #include <queue> #include <fstream> using namespace std; vector<vector<int>> tree; vector<int> ans; void BFS(int s) { queue<pair<int, int>>…

A family hierarchy is usually presented by a pedigree tree.  Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tre…

简单DFS. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<vector> using namespace std; +; vector<int>g[maxn]; int n,m; int ans[maxn]; int root; int Deep; void dfs(in…

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…

根据家谱树从根结点开始输出每一层的叶子结点数量.使用BFS来解决.因为不会重复访问结点,所以不需要vis数组来标记是否访问过该结点. //#include "stdafx.h" #include <iostream> #include <vector> #include <queue> using namespace std; vector<]; // Storing children's dynamic arrays queue<int…

统计每层的叶子节点个数建树,然后dfs即可 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <vector> using namespace std; /* 统计每层的叶子节点个数 建树,然后dfs即可 */ ; int n,m; int layer[maxn]; //统计每层的叶子节点个数 ; vector<int…

problem A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child. Input Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes i…