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            ArcTan

            dfs
            隨筆 - 16, 文章 - 117, 評論 - 6, 引用 - 0
            數據加載中……

            2008 Hangzhou 網絡賽-D hdu2421 (數論)

            Problem Description:
            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.
             

            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.
            summing the cube of number of factors of all ais.

            讀不懂題意就是傻逼啊!!!!!!!
            這個題目是要求每個因子的因子的個數然后再立方和啊啊啊啊
            8的因子有1 2 4 8,它們的因子數有1 2 3 4啊,立方和為1+8+27+64=100啊。
            轉化為算術基本定理:
            N=A^B
            求N的每個因子的因子數:
                  任何一個大于1的數可以分解成 N=a1^p1*a2^p2*a3^p3*...*an^pn, N的約數總數為(p1+1)*(p2+1)*...*(pn+1),
                  (0,1,...,p1)(0,1,...,p2)...(0,1,...,pn)
                   不難發現(1^3+2^3+...+(p1+1)^3) (1^3+2^3+...+(p2+1)^3)...(1^3+2^3+...+(pn+1)^3)即為所求。


            #include<stdio.h>
            #include
            <string.h>
            #include
            <math.h>
            #define maxn 1000005
            int p[1015];
            int  b[1015];
            int tot;

            int eular()
            {
                memset(b,
            0,sizeof(b));
                
            int i=2;tot=0;
                
            while (i<1010)
                {
                    
            while (b[i])    i++;
                    p[tot
            ++]=i;
                    
            int j=i;
                    
            while (j<1010)
                    {
                        b[j]
            =1;
                        j
            +=i;
                    }
                }
                tot
            --;
                
            return 0;
            }

            int main()
            {
                
            long long A,B;
                
            int t=0;
                eular();
                
            while (scanf("%I64d%I64d",&A,&B)==2)
                {
                    printf(
            "Case %d: ",++t);
                    B
            %=10007;
                    
            long long ans=1;
                    
            long long t,tt;
                    
            int i=0;
                    
            while (i<tot && A>1)
                    {
                        t
            =0;
                        
            while (A%p[i]==0)
                            t
            ++,A/=p[i];
                        tt
            =(t*B+1)*(t*B+2)/2 % 10007;
                        tt
            =tt*tt % 10007;
                        ans
            =(ans*tt) % 10007;
                        i
            ++;
                    }
                    
            if (A>1)
                    {
                        tt
            =(B+1)*(B+2)/2 % 10007;
                        tt
            =tt*tt % 10007;
                        ans
            =(ans*tt)%10007;
                    }
                    printf(
            "%I64d\n",ans);
                }
                
            return 0;
            }




            posted on 2012-07-19 15:09 wangs 閱讀(224) 評論(0)  編輯 收藏 引用 所屬分類: ACM-數學

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