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            posts - 14,  comments - 11,  trackbacks - 0
            Problem Description
            RSA is one of the most powerful methods to encrypt data. The RSA algorithm is described as follow:

            > choose two large prime integer p, q
            > calculate n = p × q, calculate F(n) = (p - 1) × (q - 1)
            > choose an integer e(1 < e < F(n)), making gcd(e, F(n)) = 1, e will be the public key
            > calculate d, making d × e mod F(n) = 1 mod F(n), and d will be the private key

            You can encrypt data with this method :

            C = E(m) = me mod n

            When you want to decrypt data, use this method :

            M = D(c) = cd mod n

            Here, c is an integer ASCII value of a letter of cryptograph and m is an integer ASCII value of a letter of plain text.

            Now given p, q, e and some cryptograph, your task is to "translate" the cryptograph into plain text.
             

            Input
            Each case will begin with four integers p, q, e, l followed by a line of cryptograph. The integers p, q, e, l will be in the range of 32-bit integer. The cryptograph consists of l integers separated by blanks.
             

            Output
            For each case, output the plain text in a single line. You may assume that the correct result of plain text are visual ASCII letters, you should output them as visualable letters with no blank between them.
             

            Sample Input
            101 103 7 11 7716 7746 7497 126 8486 4708 7746 623 7298 7357 3239
             

            Sample Output
            I-LOVE-ACM.
            以為要數組,想想,不是那么回事!

            挺簡單的~~
             1 #include<iostream>
             2 using namespace std;
             3 int main(){
             4     int p,q,e,t;
             5     while (cin>>p>>q>>e>>t)
             6     {
             7            int n = p*q,fn = (p-1)*(q-1);
             8            int d =1;
             9            while ((d*e)%fn!=1)d++;
            10           int c;
            11           for (int i=0;i<t;i++)
            12           {
            13                  cin>>c;
            14                  int temp=1;
            15                  for (int j=1;j<=d;j++)
            16                  {
            17                      temp*=c;
            18                      temp%=n;
            19                   }
            20                 cout<<char(temp); 
            21                  }
            22            cout<<endl;
            23            }    
            24     return 0;
            25     }
            posted on 2010-06-26 18:59 路修遠 閱讀(467) 評論(0)  編輯 收藏 引用
            <2011年4月>
            272829303112
            3456789
            10111213141516
            17181920212223
            24252627282930
            1234567

            轉載,請標明出處!謝謝~~

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            最新評論

            • 1.?re: HDU 2433 最短路
            • @test
              的確這組數據應該輸出20的
            • --YueYueZha
            • 2.?re: HDU 2433 最短路
            • 這方法應該不對。 看下面這組數據
              4 4
              1 2
              2 3
              3 4
              2 4

              畫個圖,刪去最后一條邊 2 4 后的結果應該是20,但是此方法的輸出是19
            • --test
            • 3.?re: HDU 2433 最短路
            • ans = ans + sum_u + sum_v - sum[u] - sum[v],
              這個公式不是很理解啊,不知道博主怎么想的啊,謝謝咯
            • --姜
            • 4.?re: HDU 2433 最短路
            • @attacker
              the i-th line is the new SUM after the i-th road is destroyed
            • --路修遠
            • 5.?re: HDU 2433 最短路
            • 你這樣可以AC????刪除<U,V>不僅改變 u,v最短路啊、、、求解
            • --attacker

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