【数论 && 概率论 && 打表找规律】HihoCoder

Problem Description

记 JiLi Number 指满足 前 K 个数（[1, K] ）中数字 1 的个数等于 K 的数 K 。
输出前N个中多少个JiLi Number, 同时输出最大的JiLi Number

思路：

很不错的博客 参考博客

[0, 9] 中数字1的个数 1
[0, 99] 中数字1的个数 10*2
…
[0, 999999999]中数字1的个数 10^8*9
[0, 9999999999] 中数字1的个数 10^9*10
之后可以考虑数字 1 的增长速度绝对大于数的增长速度，即后续不可能存在 JiLi Number 。

#include<bits/stdc++.h>using namespace std;#define ll long long//暴力打表ll table[85] = {1,199981,199982,199983,199984,199985,199986,199987,199988,199989,199990,200000,200001,1599981,1599982,1599983,1599984,1599985,1599986,1599987,1599988,1599989,1599990,2600000,2600001,13199998,35000000,35000001,35199981,35199982,35199983,35199984,35199985,35199986,35199987,35199988,35199989,35199990,35200000,35200001,117463825,500000000,500000001,500199981,500199982,500199983,500199984,500199985,500199986,500199987,500199988,500199989,500199990,500200000,500200001,501599981,501599982,501599983,501599984,501599985,501599986,501599987,501599988,501599989,501599990,502600000,502600001,513199998,535000000,535000001,535199981,535199982,535199983,535199984,535199985,535199986,535199987,535199988,535199989,535199990,535200000,535200001,1111111110, 1000000000000000ll};int main(){    char s[10000];    while(~scanf("%s", s))    {        int len = strlen(s);        if(len > 10) printf("83 1111111110\n");        else        {            ll sum = 0;            for(int i = 0; i < len; i++)            {                sum = sum * 10 + s[i] - '0';            }            for(int i = 0; i < 83; i++)            {                if(sum == table[i])                {                    printf("%d %lld\n", i+1, table[i]);                    break;                }                else if(sum < table[i])                {                    printf("%d %lld\n", i, table[i-1]);                    break;                }            }        }    }    return 0;}