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SCAU 13校赛 17115 ooxx numbers

17115 ooxx numbers

时间限制:1000MS  内存限制:65535K

题型: 编程题   语言: 无限制

 

Description

a number A called oo number of a number B if the sum of all As factor is the number B.
(A,B) is a pair of ooxx numbers if A is the oo number of B and B is the oo number of A.
Now i want to find how many pairs of ooxx numbers in [1,n]

输入格式

there are many cases in the input.
for each line there is a number n. ( 1 <= n <= 5000000 )

输出格式

for each n, output the numbers of pairs of ooxx numbers in [1,n]

输入样例

300
1300

输出样例

1
2

提示

hits 
220=1+2+4+71+142=284,
284=1+2+4+5+10+11+20+22+44+55+110=220。
220 and 280 is a pair of ooxx numbers.

作者

admin

 

解题思路

这题是华农13年校赛的其中一题，当初没有过，如果过了就有一等奖了，没想到这题是打表题，没有经验就想不到有这种处理方式，所以还是学到东西了，当时ZF牛打表出了点差错也没有过，回头很快的秒掉了，真的很快，这次重新写的时候还是Wa了，因为处理因子上出了点差错，将可开根的数的因子少算了一个，导致最后打出的表少了一个数据，跪了。还有一个比较傻逼的问题就是，打表就是打表，不用太在乎效率，没想到我连打表都尽量减少其运行的时间，好几个判断的条件，还是不小心折腾了一番，_(:з」∠)_

ZF大神求因子和的方法：筛素数的思维方式筛出因子和，每个i为因子，j为i的倍数，值得借鉴

相比之下自己求因子和的方法，太挫了 -_-||：

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<string>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<algorithm>
 7 
 8 using namespace std;
 9 
10 int digit[74];
11 
12 int main()
13 {
14     digit[0] = 1;
15     digit[1] = 284;
16     digit[2] = 1210;
17     digit[3] = 2924;
18     digit[4] = 5564;
19     digit[5] = 6368;
20     digit[6] = 10856;
21     digit[7] = 14595;
22     digit[8] = 18416;
23     digit[9] = 66992;
24     digit[10] = 71145;
25     digit[11] = 76084;
26     digit[12] = 87633;
27     digit[13] = 88730;
28     digit[14] = 123152;
29     digit[15] = 124155;
30     digit[16] = 139815;
31     digit[17] = 153176;
32     digit[18] = 168730;
33     digit[19] = 176336;
34     digit[20] = 180848;
35     digit[21] = 202444;
36     digit[22] = 203432;
37     digit[23] = 365084;
38     digit[24] = 389924;
39     digit[25] = 399592;
40     digit[26] = 430402;
41     digit[27] = 455344;
42     digit[28] = 486178;
43     digit[29] = 514736;
44     digit[30] = 525915;
45     digit[31] = 652664;
46     digit[32] = 669688;
47     digit[33] = 686072;
48     digit[34] = 691256;
49     digit[35] = 712216;
50     digit[36] = 783556;
51     digit[37] = 796696;
52     digit[38] = 863835;
53     digit[39] = 901424;
54     digit[40] = 980984;
55     digit[41] = 1043096;
56     digit[42] = 1099390;
57     digit[43] = 1125765;
58     digit[44] = 1189150;
59     digit[45] = 1286744;
60     digit[46] = 1292570;
61     digit[47] = 1340235;
62     digit[48] = 1438983;
63     digit[49] = 1464592;
64     digit[50] = 1483850;
65     digit[51] = 1486845;
66     digit[52] = 1598470;
67     digit[53] = 1747930;
68     digit[54] = 1749212;
69     digit[55] = 1870245;
70     digit[56] = 2062570;
71     digit[57] = 2090656;
72     digit[58] = 2429030;
73     digit[59] = 2874064;
74     digit[60] = 2928136;
75     digit[61] = 2941672;
76     digit[62] = 2947216;
77     digit[63] = 3077354;
78     digit[64] = 3716164;
79     digit[65] = 3721544;
80     digit[66] = 3892670;
81     digit[67] = 4006736;
82     digit[68] = 4300136;
83     digit[69] = 4314616;
84     digit[70] = 4445050;
85     digit[71] = 4488910;
86 
87     sort(digit, digit+72);
88     int n;
89     while(scanf("%d", &n) != EOF)
90     {
91         int i = 0;
92         for(; i<=71; ++i)
93         if(digit[i] > n) break;
94         printf("%d\n", i-1);
95     }
96     return 0;
97 }

 

posted @ 2013-07-15 14:50  Gifur  阅读(2404)  评论(0)  编辑  收藏  举报
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