hihoCoder 1432 : JiLi Number（吉利数）

hihoCoder #1432 : JiLi Number（吉利数）

时间限制:1000ms

单点时限:1000ms

内存限制:256MB

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Description - 题目描述

Driver Ji likes the digit "1". He has an accumulator which shows the sum of input number. He lists all of positive number no more than N and starts counting from one, two, three...Every time he counts a number he will add the number of digit "1" in this number to accumulator at the same time. The amazing thing happens! At some times, when he finishes counting a number X, the number which on the accumulator is X exactly, he will regard X as "JiLi Number" which means lucky number. Now he wants to know the number of "JiLi Numbers" and the biggest "JiLi Number" no more than N.

Ji司机喜欢数字””。他有一个累加器可以显示输入数的和。他列出了所有不超过N的正整数，并且开始1，，...地数了起来。每当他数到一个数时，会顺手把其数字””的个数放入累加器。不可思议的事发生了！有时他数完一个数X，累加器中的数也恰好为X，他将X称作”吉利数”，表示十分幸运的数字。现在他想知道在不超过N的情况下”吉利数”的数量与最大”吉利数”。

CN

Input - 输入

There are several test cases and the each test case is a line contains an positive integer N.(1<N≤10¹⁰⁰)

多组数据，每组数据仅有一行，每行一个正整数N。（<N≤^）

CN

Output - 输出

For each test case, output two integers which donates the number of "JiLi Numbers" and the biggest "JiLi Number".

对于每个测试用例，输出两个整数分别表示”吉利数”的数量与最大的”吉利数”。

CN

Sample Input - 样例输入

1
100000000000

Sample Output - 样例输出

1 1
83 1111111110

题解

　　暴力枚举，样例刚刚好给出了最大与最小的情况……
　　至于证明，证不出来，似乎因为是十进制数所以不能超过十位数，到了后面累加器中的值的增长速度明显超过了数数速度，应该是没有交集了……

代码 C++

 #include<cstdio>
 #include <algorithm>
 #define ll long long
 #define mx 84
 ll opt[mx] = {
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     0x7FFFFFFFFFFFFFFF
 };
 int main(){
     char rd[];
     int i;
     ll tmp;
     while (gets(rd)){
         tmp = ;
         for (i = ; i <  && rd[i]; ++i) tmp = tmp *  + rd[i] - '';
         i = std::upper_bound(opt, opt + mx, tmp) - opt;
         printf("%d %lld\n", i, opt[i - ]);
     }
     return ;
 }