poj 1019 Number Sequence(dp+二分)

Number Sequence

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 33439 Accepted: 9559

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

83

Sample Output

22

Source

Tehran 2002, First Iran Nationwide Internet Programming Contest

链接：http://poj.org/problem?id=1019

【题意】数列：112123123412345123456123456712345678123456789123456789101234567891011123456789101112......

求第i位的数字

【思路】

1.打表。求出子数列1-数x(1,2,3...)包含数字的个数，求出整个数列1-数x（1,2,3...）包含数字的个数；

2.根据  i  求出i所在的子数列（1-x）,再求出第i位所在的数y,再求出数字ans;

【代码】

#include<cstdio>#include<iostream>using namespace std;long long dp[100000];//测出最大是31268long long sum[100000];int r;int get(long long x){    int ans=0;    while(x)    {        ++ans;        x=x/10;    }    return ans;}void innit(){    int i;    dp[0]=0;    sum[0]=0;    for(int i=1;1;i++)    {        dp[i]=dp[i-1]+get(i);        sum[i]=sum[i-1]+dp[i];        //if(i==9||i==99||i==999||i==9999||i==99999||i==999999)            //cout<<sum[i]<<ends;        if(sum[i]>=2147483647)        {            r=i;            //cout<<r;            break;        }    }}int main(){    innit();    int n,c;    scanf("%d",&c);    while(c--)    {        scanf("%d",&n);        int left=1,right=r;        while(left<right)        {            int mid=(left+right)/2;            if(sum[mid]<n)                left=mid+1;            else                right=mid;        }        n=n-sum[left-1];        right=left;        left=1;        while(left<right)        {            int mid=(left+right)/2;            if(dp[mid]<n)                left=mid+1;            else                right=mid;        }        n=n-dp[left-1];        int num[10],lenth=dp[left]-dp[left-1];        for(int i=1;i<=lenth;i++)        {            num[lenth-i+1]=left%10;            left=left/10;        }        cout<<num[n]<<endl;    }    return 0;}