NYOJ 413 月赛的悲剧

话说这道题是月赛时的第一题，想这道题至少想了一个半小时，当时的基本思路已经想出来了，就有一个地方一直实现不了，于是就一直在想，悲剧的是，最后还是没能实现。更悲剧的是，最后一个半小时，一次也没提交。。昨天又想了想，还是卡在了那个地方，刚才突然灵光一现，，，，想了出来，，不容易啊。。。。

题目：

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

输入

The first line of the input file contains a single integer t (1 ≤ t ≤ 20), 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)

输出

There should be one output line per test case containing the digit located in the position i.

样例输入

2
8
3

样例输出

2
2

ac代码：

#include <iostream> #include <string.h> #include <cstdio> using namespace std; long long sum[35000],a[35000]; void chushihua(){ sum[1]=1;a[1]=1; for(int i=2;i<=9;++i){ sum[i]=sum[i-1]+a[1]+(i-1); a[i]=sum[i]-sum[i-1]; } for(int i=10;i<=99;++i){ sum[i]=sum[i-1]+a[9]+(i-9)*2; a[i]=sum[i]-sum[i-1]; } for(int i=100;i<=999;++i){ sum[i]=sum[i-1]+a[99]+(i-99)*3; a[i]=sum[i]-sum[i-1]; } for(int i=1000;i<=9999;++i){ sum[i]=sum[i-1]+a[999]+(i-999)*4; a[i]=sum[i]-sum[i-1]; } for(int i=10000;i<=34999;++i){ sum[i]=sum[i-1]+a[9999]+(i-9999)*5; a[i]=sum[i]-sum[i-1]; } } int main(){ int k; chushihua(); scanf("%d",&k); while(k--){ long long n; cin>>n; long long kk; for(int i=1;i<=34999;++i){ if(sum[i]<n&&sum[i+1]>=n) { kk=i;break; } } n-=sum[kk]; int x=1; for(int i=0;i<34999;++i){ if(a[i]<n&&a[i+1]>=n) {x=i;break;} } n-=a[x]; x++; int b[11],zz=1; while(x){ b[zz++]=x%10; x/=10; } cout<<b[zz-n]<<endl; } return 0; }