UVA - 10706 Number Sequence（贪心）

Problem B
Number Sequence
Input: standard input
Output: standard output
Time Limit: 1 second

A single positive integer iis given. Write a program to find the digit located in the position iin the sequence of number groups S₁S₂�S_k. Each group S_kconsists 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 <=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

题目大意：给你一个升序的序列，序列由如下序列组成
1
12
123
1234
12345
123456
1234567
12345678
123456789
12345678910
1234567891011
12345678910
现在给你一个数字n，代表在第几个位置，现在请你输出这个位置的数字。
(1 <=n <=2147483647)

解析：这题主要用到贪心的思想。

先打表，开一个数组digit保存每行的数字的个数，再开一个数组sum保存到当前行所有数字的个数，因为n的值可能很大，sum要开成long long。

现在遍历sum的数组，找到第一个sum[i] >= n，计算出sum[i-1] 到 n的距离dis，再用一个cnt来计算到sum[i-1]的距离，当cnt >= dis时，求出cnt - dis 的距离，这个距离就是当前该数字的第几位，输出该位数的值。

#include <cstdio>
#include <cstring>
#include <cmath>
#define ll long long
using namespace std;
const int N = 100000;
ll sum[N];
ll digit[N];
void init() {
	sum[0] = 0;
	digit[0] = 0;
	for(int i = 1; i < N; i++) {
		digit[i] = digit[i-1] + log10(i) + 1;
		sum[i] = sum[i-1] + digit[i];
	}
}
int solve(int n) {
	for(int i = 1; i < N; i++) {
		if(n <= sum[i]) {
			ll dis = n - sum[i-1];
			int ans,cnt = 0;
			int j;
			for(j = 1; j <= i; j++) {
				cnt += (log10(j) + 1);
				if(cnt >= dis) {
					break;
				}
			}
			if(cnt == dis) {
				ans = j % 10;
				return ans;
			}else if(cnt > dis) {
				ans = j / pow(10,(cnt-dis));
				ans %= 10;
				return ans;
			}
		}
	}
	return 0;
}
int main() {
	init();
	int t,n;
	scanf("%d",&t);
	while(t--) {
		scanf("%d",&n);
		int ans = solve(n);
		printf("%d\n",ans);
	}
	return 0;
}