我是在做1019题时遇到问题的。计算机判断时给我的结果是“Exceed Time Limit”，但我不觉得这个程序能执行1秒钟还完不成。但又不好测试（还不会仿真往计算机里输入和读取数据）。程序本身的结果经过测试应该是没问题的。希望好心的诸位提提意见，看看我哪里做的不对头？ 谢谢！！
题目是：
Number Sequence

Time Limit:1000MS  Memory Limit:10000K
Total Submit:1433 Accepted:322 

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

Source 
First Iran Nationwide Internet Programming Contest Problem D
===================================================================
我给出了程序：
#include <iostream>
using namespace std;

int GetDigitByPosition(int nPosition)
{
	int i, nMax, nLength, nFactor, nDigit;
	nMax = 1;
	nFactor = 1000000;
	while(true)
	{
		for(i = 1; i <= nMax; i++)
		{
			if(i < 10)
			{
				nLength = 1;
			}
			else if(i < 100)
			{
				nLength = 2;
			}
			else if(i < 1000)
			{
				nLength = 3;
			}
			else if(i < 10000)
			{
				nLength = 4;
			}
			else if(i < 100000)
			{
				nLength = 5;
			}
			else if(i < 1000000)
			{
				nLength = 6;
			}
			else
			{
				return -1;
			}

			if((nPosition - nLength) <= 0)
			{
				goto find;
			}
			else
			{
				nPosition -= nLength;
			}
		}
		nMax++;
	}
find:
	while((i / nFactor) == 0)
	{
		nFactor /= 10;
	}

	while(true)
	{
		nDigit = i / nFactor;

		if(nPosition == 1)
		{
			return nDigit;
		}
		else
		{
			nPosition--;
			i %= nFactor;
			nFactor /= 10;
		}

	}
}
int main()
{
	int nNum = 0;
	int i, nPosition[10], nDigit[10];
	cin>>nNum; //load the number of test cases
	for(i = 0; i < nNum; i++)
	{
		cin>>nPosition[i];
		nDigit[i] = GetDigitByPosition(nPosition[i]);
	}
	
	//output the result
	for(i = 0; i < nNum; i++)
	{
		cout<<nDigit[i]<<endl;
	}
	return 0;
}

• 不要过分自信，你的程序绝对绝对会超时 (2710) frkstyc 2005-08-12 00:02:21
  • Re:不要过分自信，你的程序绝对绝对会超时 (5) speedfirst 2005-08-12 00:03:50
    • 运行你的程序，输入1，回车，输入1000000000，回车 (9) frkstyc 2005-08-12 00:05:11
      • Re:运行你的程序，输入1，回车，输入1000000000，回车 (24) speedfirst 2005-08-12 00:07:24
  • Re:不要过分自信，你的程序绝对绝对会超时 (35) speedfirst 2005-08-12 10:34:23
    • Re:不要过分自信，你的程序绝对绝对会超时 (25) speedfirst 2005-08-12 11:06:18
      • done，解决了 (31) speedfirst 2005-08-12 11:24:59
      • Re:不要过分自信，你的程序绝对绝对会超时 (44) fyliang_wh 2006-04-05 20:11:51