您的位置:首页 > 产品设计 > UI/UE

UVA 10706 Number Sequence (找规律 + 打表 + 查找)

2013-08-22 18:50 337 查看
Problem B
Number Sequence
Input:
standard input
Output: standard output
Time Limit: 1 second

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 80digits of the sequence are as follows:

11212312341234512345612345671234567812345678912345678910123456789101112345678910

Input

The first line of the input file contains a single integer t (1 <=t <=25), 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 Output for Sample Input

2

8

3

2

2

题意: 给定一个数字n,要求出第n位对应题目给定那个序列的数字是多少,序列为1 , 1, 2, 1, 2, 3, 1, 2, 3, 4.。。。以此类推。注意2位数是占2个数,3位数是占3个数这样的。

思路:题目数据为2^31 - 1。直接暴力肯定跪。所以可以分区间讨论,个位数,十位数,百位数,这样去讨论,个位数有9个区间1,[1,2],[1,3]...[1,9]一共45位。以此类推。可以位数对应的第几个区间。打出一个表来。然后根据位数在区间去查找。最后找出对应位数上的数字。

代码:

#include <stdio.h>
#include <string.h>
#include <math.h>
int t, n, pos;
long long num[100005];

void init() {//我去年打了个表
pos = 0;
memset(num, 0, sizeof(num));
for (int i = 1; i <= 5; i ++) {
for (long long j = pow(10, i - 1); j < pow(10, i); j ++) {
num[j] = num[j - 1];
int d = (j - pow(10, i - 1) + 1) * i;
for (long long k = 1; k < i; k ++)
d += (pow(10, k) - pow(10, k - 1)) * k;
num[j] += d;
}
}
}

void find() {//找出对应区间
for (int i = 1; i < 100005; i ++)
if (n > num[i - 1] && n <= num[i]) {
pos = i;
break;
}
}

int solve() {//找出对应数字。
int wei = n - num[pos - 1];
int sum = 0;
for (int i = 1; i <= pos; i ++) {
int sb = i;
int j;
for (j = 5; j >= 0; j --)
if (sb >= int(pow(10, j)))
break;
for (; j >= 0; j --) {
if (sb >= int(pow(10, j))) {
int fuck = sb / int(pow(10, j));
sb -= fuck * int(pow(10, j));
sum ++;
if (sum == wei)
return fuck;
}
else {
sum ++;
if (sum == wei)
return 0;
}
}
}
}
int main() {
init();
scanf("%d", &t);
while (t --) {
scanf("%d", &n);
find();
printf("%d\n", solve());
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: