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 example, given N being 12, there are five 1's in 1, 10, 11, and 12.

Input Specification:

Each input file contains one test case which gives the positive N (<=2³⁰).

Output Specification:

For each test case, print the number of 1's in one line.

Sample Input:

Sample Output:

提交代码

思路：

统计每位的1的贡献。

对于k位（k>=1）：

1.Ak=0，count+=AnAn-1....Ak+1AkAk-1....A1*10^(k-1)

2.Ak=1，count+=AnAn-1....Ak+1AkAk-1....A1*10^(k-1)+Ak-1Ak-2...A1+1

3.Ak>=2，count+=(AnAn-1....Ak+1AkAk-1....A1+1)*10^(k-1)

 1 #include<cstdio>
 2 #include<stack>
 3 #include<cstring>
 4 #include<iostream>
 5 #include<stack>
 6 #include<set>
 7 #include<map>
 8 using namespace std;
 9 //count的最大值是1036019223
10 int main(){
11     int n;
12     scanf("%d",&n);
13     long long base=1;
14     long long count=0;
15     int frpart,afpart,a;
16     while(n>=base){
17         a=n/base%10;
18         frpart=n/(10*base);
19         afpart=n%base;
20         count+=frpart*base;
21         if(a==1){
22             count+=afpart+1;
23         }
24         else if(a>1){
25             count+=base;
26         }
27         base*=10;
28     }
29     printf("%lld\n",count);
30     return 0;
31 }

转载于:https://www.cnblogs.com/Deribs4/p/4776672.html

pat1049. Counting Ones (30)相关推荐

pat1004. Counting Leaves (30)
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 fam ...
PAT甲级1049 Counting Ones (30 分)：[C++题解]统计1的个数、数位统计
文章目录题目分析题目链接题目分析来源:PAT网站分析: 以数字abcdefg这个7位数字为例,说一下本题的思路. 1)数字1在每一位出现的次数. 2)以第d位为例,第d位的取值可以分为3种情 ...
PAT甲级1004 Counting Leaves (30分)：[C++题解]树、邻接表存储树、dfs遍历树
文章目录题目分析题目链接题目分析题意重述:一棵树,求每一层的叶子节点数目. 分析构造树,使用邻接表来存(相当于存储有向图). 需要一个头结点数组h[N],然后每个头节点往外形成一个单链表e[ ...
1004. Counting Leaves (30)
时间限制 400 ms 内存限制 32000 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue A family hierarchy is usually pr ...
1004 Counting Leaves (30分) (vector实现)
题解: 用vector邻接表建图,两个邻接表实现层序遍历. 有注释在代码代码: /*Keep on going Never give up*/ #pragma GCC optimize(3,&quo ...
1049. Counting Ones (30)
题目如下: The task is simple: given any positive integer N, you are supposed to count the total number o ...
1049 Counting Ones (30 分)【难度: 难 / 知识点: 分治 / DP】
https://pintia.cn/problem-sets/994805342720868352/problems/994805430595731456 方法一: 找规律,分治做法. //0-999 ...
1004 Counting Leaves (30 分)【难度: 中 / 知识点: 树的遍历】
题目意思: 求每一层的叶子结点数方法一: 深搜数组模拟存储邻接表 #include<bits/stdc++.h> using namespace std; const int N=1e ...
【两种解法】1004 Counting Leaves (30 分)_27行代码AC
立志用最少的代码做最高效的表达 PAT甲级最优题解-->传送门 A family hierarchy is usually presented by a pedigree tree. Your ...

pat1049. Counting Ones (30)

1049. Counting Ones (30)

pat1049. Counting Ones (30)相关推荐

最新文章

热门文章