POJ 1019 Number Sequence

题意：问有如下规律的数列的第n项是啥：

11212312341234512345612345671234567812345678912345678910123456789101112345678910

解法：每段连续数字为一组，算每组序列的长度，累加直到超过n，说明n在前一组连续数字内，枚举组内数字，累加长度直到超过n，说明n在前一数字中，找出这一数位输出。

总的来说就是一有点恶心的模拟……最后我想说……

注意longlong

对……每次都记不住……_(:з」∠)_

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<string>
#include<string.h>
#include<math.h>
#include<limits.h>
#include<time.h>
#include<stdlib.h>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#define LL long long

using namespace std;

int main()//变量名起的变幻莫测不忍吐槽嗷
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        LL n;
        scanf("%I64d", &n);
        LL sum = 0, x = 0, i = 1;
        for(; n > sum; i++)
        {
            x += (LL)log10(i + 0.5) + 1;
            sum += x;
        }
        i--;
        LL tmp = n - (sum - x);
        LL sum1 = 0;
        for(LL j = 1; j <= i; j++)
        {
            sum1 += (LL)log10(j + 0.5) + 1;
            if(sum1 >= tmp)
            {
                if(sum1 > tmp)
                    sum1 -= (LL)log10(j + 0.5) + 1;
                LL ttmp = tmp - sum1;
                vector <int> ans;
                LL x1 = j;
                while(x1)
                {
                    ans.push_back(x1 % 10);
                    x1 /= 10;
                }
                if(ttmp == 0) ttmp = ans.size();
                printf("%d
", ans[ans.size() - ttmp]);
                break;
            }
        }
    }
    return 0;
}