Xiaoming has just come up with a new way for encryption, by calculating the key from a publicly viewable number in the following way:
Let the public key N = AB, where 1 <= A, B <= 1000000, and a0, a1, a2, …, ak-1 be the factors of N, then the private key M is calculated by summing the cube of number of factors of all ais. For example, if A is 2 and B is 3, then N = AB = 8, a0 = 1, a1 = 2, a2 = 4, a3 = 8, so the value of M is 1 + 8 + 27 + 64 = 100.
However, contrary to what Xiaoming believes, this encryption scheme is extremely vulnerable. Can you write a program to prove it?
Input
There are multiple test cases in the input file. Each test case starts with two integers A, and B. (1 <= A, B <= 1000000). Input ends with End-of-File.
Note: There are about 50000 test cases in the input file. Please optimize your algorithm to ensure that it can finish within the given time limit.
Output
For each test case, output the value of M (mod 10007) in the format as indicated in the sample output.
Sample Input
2 2
1 1
4 7
Sample Output
Case 1: 36
Case 2: 1
Case 3: 4393
Author: 2008 Asia Hangzhou Regional Contest Online
這道題是杭州賽區的網絡預選賽的賽題，當時打死也沒過，現在也沒有當時的代碼了，不知道哪里錯了，今天一看想起了一個公式就是
1^3+2^3+……n^3=(n)^2*(n+1)*^2/4;記得這個公式還是高二老師教會的著，具體證明我們可以數學歸納法證明一下，簡單的很自己證明吧！
而那時自己網絡賽的時候不知道怎么推導出過10007是個循環，可以把b的取模，而現在有了這個公式就可以一步搞定為什么對b取模！
要解題就必須求出a ^b的因子數，這不是難點
證明如下：
N=a1^b1*a2^b2*……*aN^bN;根據排列組合原理我們知道N的因子數可以如下解出：
即(b1+1)*(b2+1)*.......*(bN+1);而現在是a^b只需要把(b1*b+1)*(b2*b+1)*.......*(bN*b+1);極為因子數：
而這個時候我們知道1<=a,b<=1000000超出整形范圍：我們進一步優化得到
因為有：m*n%x=（m%x*n%x）;（m+n）%x=（m%x+n%x）;
所以我們必須在過程中優化算法！
這樣這道題目就出來了！當然求因子時必不可少素數打表！可以優化打表1000，減少時間！當然還可以把1-10007的密碼表打出，更優時間啊
素數打表代碼如下：
memset(prim,0,sizeof(prim));
????for(sign=0,i=2;i<MAXN;i++)
????????if(!prim[i])
{
????????????num[sign++]=i;
????????????for(j=0;i*j<MAXN;j++)
????????????????prim[i*j]=true;????
????????}
posted on 2009-03-23 19:20
KNIGHT 閱讀(537)
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