Android京东笔试 算法第二题
/* 基于上一篇帖子中别人给的解法的改进，不知道这种遍历会不会通过 */
/*
小明同学学习了不同的进制之后，拿起了一些数字做起了游戏。小明同学知道，在日常生活中我们最常用的是十进制数，而在计算机中，二进制数也很常用。现在对于一个数字x，小明同学定义出了两个函数f(x)和g(x)。
f(x)表示把x这个数用十进制写出后各个数位上的数字之和。如f(123)=1+2+3=6。
g(x)表示把x这个数用二进制写出后各个数位上的数字之和。如123的二进制表示为1111011，那么g(123)=1+1+1+1+0+1+1=6。
小明同学发现对于一些正整数x满足f(x)=g(x)，他把这种数字称为幸运数，现在他想知道，小于等于n的幸运数有多少个。
输入
第一行一个整数T(T<=10000)表示数据组数，每组数据输入一个数n(n<=100000)。
输出
每组数据输出一行，小于等于n的幸运数个数。
样例输入
3
1
5
21
样例输出
1
1
3
Hint
小于等于21的三个幸运数分别为：1,20,21。
*/
#include<stdio.h>
#define MAX 483
int main(){
int N;
int arr[483] =
{1, 20, 21, 122, 123, 202, 203, 222, 223, 230, 231, 302,
303, 410, 411, 502, 503,
1130, 1131, 1150, 1151, 1202, 1203, 1212, 1213, 1230,
1231, 1300, 1301, 1402, 1403, 1502,
1503, 1510, 1511, 2006, 2007, 2032, 2033, 2102, 2103,
2200, 2201, 3006,
3007, 3012, 3013, 3050, 3051, 3070, 3071, 3102, 3103,
3110, 3111, 3300, 3301,
4014, 4015, 5000, 5001, 5010, 5011, 5020, 5021, 6002,
6003, 6012, 6013, 6102,
6103, 7000, 7001, 7102, 7103, 8110, 8111, 10006, 10007,
10032, 10033, 10050, 10051,
10060, 10061, 10070, 10071, 10112, 10113, 10132, 10133,
10170, 10171, 10234,
10235, 10310, 10311, 10410, 10411, 10430, 10431, 11012,
11013, 11050, 11051, 11070,
11071, 11114, 11115, 11134, 11135, 11150, 11151, 11214,
11215, 11240, 11241,
11250, 11251, 11260, 11261, 11300, 11301, 11502, 11503,
11710, 11711, 12006, 12007,
12024, 12025, 12112, 12113, 12142, 12143, 12204, 12205,
12214, 12215, 12250
, 12251, 13012, 13013, 13050, 13051, 13200, 13201, 13230,
13231, 14014, 14015, 14020,
14021, 14102, 14103, 14330, 14331, 15002, 15003, 15010,
15011, 15020, 15021,
15030, 15031, 15210, 15211, 15230, 15231, 15310, 15311,
16010, 16011, 16030,
16031, 16102, 16103, 16120, 16121, 16300, 16301, 16310,
16311, 20004, 20005, 20040,
20041, 20050, 20051, 20060, 20061, 20112, 20113, 20142,
20143, 20206, 20207,
20230, 20231, 20312, 20313, 20320, 20321, 20340, 20341,
20414, 20415, 20420, 20421,
21002, 21003, 21022, 21023, 21030, 21031, 21112, 21113,
21202, 21203, 21212
, 21213, 21220, 21221, 21302, 21303, 21400, 21401, 22006,
22007, 22030, 22031, 22120,
22121, 22130, 22131, 22140, 22141, 22210, 22211, 22220,
22221, 22230, 22231,
22302, 22303, 22310, 22311, 22500, 22501, 23004, 23005,
23012, 23013, 23110,
23111, 23210, 23211, 23230, 23231, 24002, 24003, 24012,
24013, 24022, 24023, 24100,
24101, 24302, 24303, 24510, 24511, 26100, 26101, 30006,
30007, 30042, 30043,
30050, 30051, 30060, 30061, 30070, 30071, 30104, 30105,
30112, 30113, 30132, 30133,
30150, 30151, 30210, 30211, 30220, 30221, 30230, 30231,
30320, 30321, 30400
, 30401, 30420, 30421, 30500, 30501, 30702, 30703, 31004,
31005, 31012, 31013, 31120,
31121, 31150, 31151, 31214, 31215, 31300, 31301, 31400,
31401, 31410, 31411,
31420, 31421, 31610, 31611, 31710, 31711, 32002, 32003,
32012, 32013, 32022,
32023, 32112, 32113, 32212, 32213, 32250, 32251, 32302,
32303, 32410, 32411, 32502,
32503, 32700, 32701, 40002, 40003, 40012, 40013, 40022,
40023, 40122, 40123,
40200, 40201, 40210, 40211, 40220, 40221, 40400, 40401,
40430, 40431, 40510, 40511,
40700, 40701, 41210, 41211, 42100, 42101, 43004, 43005,
44000, 44001, 44020
, 44021, 45000, 45001, 45010, 45011, 45020, 45021, 47100,
47101, 50002, 50003, 50012,
50013, 50020, 50021, 50300, 50301, 51010, 51011, 51020,
51021, 51030, 51031,
51130, 51131, 52010, 52011, 52030, 52031, 52110, 52111,
52200, 52201, 52210,
52211, 52220, 52221, 53102, 53103, 53210, 53211, 54010,
54011, 60002, 60003, 60012,
60013, 60022, 60023, 60102, 60103, 60120, 60121, 60200,
60201, 60210, 60211,
60220, 60221, 60300, 60301, 60310, 60311, 60400, 60401,
61002, 61003, 61022, 61023,
61030, 61031, 61210, 61211, 61302, 61303, 61400, 61401,
62010, 62011, 62110
, 62111, 62200, 62201, 63102, 63103, 65000, 65001, 65010,
65011, 65020, 65021, 70010,
70011, 70110, 70111, 71100, 71101, 80110, 80111, 90000,
90001, 90030, 90031 };
scanf("%d",&N);
while(N--){
int n;
int i=0;
int flag = 0;
scanf("%d",&n);
while(n && flag ==0){
for(i=0;i<MAX;i++){
if(n == arr[i]){
flag = 1;
printf("%d\n",i+1);
}
}
n--;
}
}
return 0;
}
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