poj 1019 Number Sequence
2016-08-23 19:05
435 查看
题目链接:点击打开链接
Description
A single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2...Sk. Each group Sk consists of a sequence of positive integer numbers ranging from 1 to k, written one after another.
For example, the first 80 digits of the sequence are as follows:
11212312341234512345612345671234567812345678912345678910123456789101112345678910
Input
The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by one line for each test case. The line for a test case contains the single integer i (1 ≤ i ≤ 2147483647)
Output
There should be one output line per test case containing the digit located in the position i.
Sample Input
Sample Output
Description
A single positive integer i is given. Write a program to find the digit located in the position i in the sequence of number groups S1S2...Sk. Each group Sk consists of a sequence of positive integer numbers ranging from 1 to k, written one after another.
For example, the first 80 digits of the sequence are as follows:
11212312341234512345612345671234567812345678912345678910123456789101112345678910
Input
The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by one line for each test case. The line for a test case contains the single integer i (1 ≤ i ≤ 2147483647)
Output
There should be one output line per test case containing the digit located in the position i.
Sample Input
2 8 3
Sample Output
2 2
#include <iostream> #define MAX 31269 #include<cmath> using namespace std; int a[MAX]; long long b[MAX]; int main() { a[1]=1; b[1]=1; for(int i=2; i<MAX; i++ ) { a[i]=a[i-1]+ (int )log10((double)i)+1; b[i]=b[i-1]+a[i]; //log10(i)+1 表示第i组数字列的长度 比 第i-1组 长的位数 //前i组的长度s[i] 等于 前i-1组的长度s[i-1] + 第i组的长度a[i] } //log()是重载函数,必须对int的i强制类型转换,以确定参数类型 int n; int t; cin>>t; while(t--) { cin>>n; int i=1; while(b[i]<n) { i++; } int pos=n-b[i-1]; int len=0; for(i=1; len<pos; i++) { len+=(int )log10((double)i)+1; } cout<<(i-1)/(int )pow((double)10,len-pos)%10<<endl; //之所以i-1,是因为前面寻找第i组长度时,i++多执行了一次 //i=i-1 此时i刚好等于第n位个置上的数 (数是整体,例如123一百二十三,i刚好等于123,但n指向的可能是1,2或3) //pos为n指向的数字在第i组中的下标值 //len为第i组的长度 //那么len-pos就是第i组中pos位置后多余的数字位数 //则若要取出pos位上的数字,就要利用(i-1)/pow(10,len-pos)先删除pos后多余的数字 //再对剩下的数字取模,就可以得到pos //例如要取出1234的2,那么多余的位数有2位:34。那么用1234 / 10^2,得到12,再对12取模10,就得到2 } return 0; }
相关文章推荐
- Kylin Cube Build流程
- 交叉编译参数--build、host和target的区别
- arduino入门套件学习过程-安装配置
- NSInteger 与 NSUInteger 区别
- UIAutomator - Hello World
- iOS本地通知 UILocalNotification
- pcDuino day1
- 队列同步器(AQS)详解
- Android02--Android之SpannableString +SpannableStringBuilder
- Looping Techniques
- 300. Longest Increasing Subsequence
- Druid连接池简单入门配置
- android Fragment 更新某个UI
- UNREAL ENGINE 4 BUILD FILE DEMYSTIFIED
- Unreal Engine 4 —— 禅意花园项目中的水池
- 编译caffe出现错误:make: *** [.build_release/src/caffe/common.o] Error 1
- Vue 资源
- StringBuffer,and,StringBuilder
- NYOJ 714 Card Trick(队列queue)
- 解决PKIX:unable to find valid certification path to requested target 的问题