hihoCoder #1432 : JiLi Number（吉利数）
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”1”。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1，2，3...地数了起来。每当他数到一个数时，会顺手把其数字”1”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”。
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。（1<N≤10^100）
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”。
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++
1 #include<cstdio> 2 #include <algorithm> 3 #define ll long long 4 #define mx 84 5 ll opt[mx] = { 6 1, 7 199981, 8 199982, 9 199983, 10 199984, 11 199985, 12 199986, 13 199987, 14 199988, 15 199989, 16 199990, 17 200000, 18 200001, 19 1599981, 20 1599982, 21 1599983, 22 1599984, 23 1599985, 24 1599986, 25 1599987, 26 1599988, 27 1599989, 28 1599990, 29 2600000, 30 2600001, 31 13199998, 32 35000000, 33 35000001, 34 35199981, 35 35199982, 36 35199983, 37 35199984, 38 35199985, 39 35199986, 40 35199987, 41 35199988, 42 35199989, 43 35199990, 44 35200000, 45 35200001, 46 117463825, 47 500000000, 48 500000001, 49 500199981, 50 500199982, 51 500199983, 52 500199984, 53 500199985, 54 500199986, 55 500199987, 56 500199988, 57 500199989, 58 500199990, 59 500200000, 60 500200001, 61 501599981, 62 501599982, 63 501599983, 64 501599984, 65 501599985, 66 501599986, 67 501599987, 68 501599988, 69 501599989, 70 501599990, 71 502600000, 72 502600001, 73 513199998, 74 535000000, 75 535000001, 76 535199981, 77 535199982, 78 535199983, 79 535199984, 80 535199985, 81 535199986, 82 535199987, 83 535199988, 84 535199989, 85 535199990, 86 535200000, 87 535200001, 88 1111111110, 89 0x7FFFFFFFFFFFFFFF 90 }; 91 int main(){ 92 char rd[200]; 93 int i; 94 ll tmp; 95 while (gets(rd)){ 96 tmp = 0; 97 for (i = 0; i < 15 && rd[i]; ++i) tmp = tmp * 10 + rd[i] - '0'; 98 i = std::upper_bound(opt, opt + mx, tmp) - opt; 99 printf("%d %lld ", i, opt[i - 1]); 100 } 101 return 0; 102 }