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…

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…

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…

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…

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…

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…

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties: The left subtree of a node contains only nodes with keys less than or equal to the node's key. The right subtree of a node contains only nodes with…

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…