PAT 1049 Counting Ones
2013-08-05 15:25
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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.
本题的题意也是十分明确的,就是给定一个小于等于2^30的正整数,要求计算出1~n中,1出现的次数。
小编一开始也摸不着头脑,之后呢,通过猜想假设,1出现的次数存在一定特性,先遍历出些小规模的数据,
最后得到了一个结论:
设n由abcdefg组成,1的个数由sum来计数,若a=1,sum+=bcdefg+1+(1~bcdefg中1出现的次数);
否则sum+=n-bcdefg+a*(n-1位数中1出现的总次数)+(1~bcdefg中1出现的次数)。
参考代码如下:
总结:这种题目多数运用的是数论相关知识,小编就是缺乏对数论的了解,在做本题时用时过长,数论知识的扩充是必不可少的。有空得多看看相关书籍了=.=...
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.
本题的题意也是十分明确的,就是给定一个小于等于2^30的正整数,要求计算出1~n中,1出现的次数。
小编一开始也摸不着头脑,之后呢,通过猜想假设,1出现的次数存在一定特性,先遍历出些小规模的数据,
最后得到了一个结论:
设n由abcdefg组成,1的个数由sum来计数,若a=1,sum+=bcdefg+1+(1~bcdefg中1出现的次数);
否则sum+=n-bcdefg+a*(n-1位数中1出现的总次数)+(1~bcdefg中1出现的次数)。
参考代码如下:
#include<stdio.h> #include<math.h> int sum(int n,int a[]) { int num=0,count=0,x,y=n;; while(y){count++;y/=10;} if(count==1)return 1;//当n为个位数时返回次数1 x=(int)pow(10.0,count-1); num+=(n/x)*a[count-1]; if(n/x==1) num+=n%x+1; else num+=x; num+=sum(n%x,a);//进行递归计数 return num; } int main() { int n,number,a[20]; int i,x; for(i=1,x=1;i<10;i++) { a[i]=i*x; x*=10; }//i位数中总的1出现次数为a[i] while(scanf("%d",&n)) { x=n; number=sum(n,a);//计算1出现次数总和 printf("%d\n",number); } return 0; }
总结:这种题目多数运用的是数论相关知识,小编就是缺乏对数论的了解,在做本题时用时过长,数论知识的扩充是必不可少的。有空得多看看相关书籍了=.=...
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