BZOJ 4454: C Language Practice

4454: C Language Practice

Time Limit: 20 Sec  Memory Limit: 24 MB
Submit: 501  Solved: 112
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Description

Input

第一行输入一个正整数T(T<=85)，表示测试数据的组数。

每组数据第一行包含两个正整数n,m(1<=n,m<=2000)，表示序列的长度。

第二行包含n个正整数，表示a[0],a[1],...,a[n-1](0<=a[i]<=1000000)。

第三行包含m个正整数，表示b[0],b[1],...,b[m-1](0<=b[i]<=1000000)。

Output

对于每组数据输出一行一个整数，即答案。

Sample Input

3
3 2
5 9 6
3 4
2 2
8 9
0 6
1 1
9
6

Sample Output

6
22
3

HINT

注意：此题只有一个数据点。

Source

 

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$O(N)-O(1)$的gcd黑科技

 #include <cstdio>

 inline char Char(void)
 {
     static const int siz = ;

     static char buf[siz];
     static char *hd = buf + siz;
     static char *tl = buf + siz;

     if (hd == tl)
         fread(hd = buf, , siz, stdin);

     return *hd++;
 }

 inline int Int(void)
 {
     int ret = , neg = , c = Char();

     for (; c < ; c = Char())
         if (c == '-')neg ^= true;

     for (; c > ; c = Char())
         ret = ret *  + c - '';

     return neg ? -ret : ret;
 }

 int g[][];

 int fac[][];

 int pre[], pri[], tot;

 inline int gcd(int a, int b)
 {
     if (a <=  && b <= )
         return g[a][b];

     int ret = ;

     for (int i = ; i < ; ++i)
         if (fac[a][i] != )
         {
             int t = fac[a][i];

             if (pre[t])
             {
                 int d = g[t][b % t];
                 ret *= d, b /= d;
             }
             else if (b % t == )
                 ret *= t, b /= t;
         }

     return ret;
 }

 signed main(void)
 {
     for (int i = ; i <= ; ++i)
         for (int j = ; j <= ; ++j)
             if (i ==  || j == )
                 g[i][j] = i + j;
             else if (g[j][i % j])
                 g[i][j] = g[j][i % j];
             else if (g[i % j][j])
                 g[i][j] = g[i % j][j];
             else if (g[i][j % i])
                 g[i][j] = g[i][j % i];
             else if (g[j % i][i])
                 g[i][j] = g[j % i][i];

     fac[][] = fac[][] = fac[][] = ;

     for (int i = ; i <= ; ++i)
     {
         if (!pre[i])pri[++tot] = i,
             fac[i][] = fac[i][] = , fac[i][] = i;

         for (int j = ; j <= tot && pri[j] * i <= ; ++j)
         {
             int num = pri[j] * i, p = pri[j];

             pre[num] = p;
             fac[num][] = fac[i][];
             fac[num][] = fac[i][];
             fac[num][] = fac[i][];

             if (fac[num][] * p <= )
                 fac[num][] *= p;
             else if (fac[num][] * p <= )
                 fac[num][] *= p;
             else
                 fac[num][] *= p;
         }
     }

     for (int cas = Int(); cas--; )
     {
         static int n, m, a[], b[];

         n = Int();
         m = Int();

         for (int i = ; i < n; ++i)
             a[i] = Int();

         for (int i = ; i < m; ++i)
             b[i] = Int();

         unsigned ans = ;

         for (int i = ; i < n; ++i)
             for (int j = ; j < m; ++j)
                 ans += gcd(a[i], b[j]) ^ i ^ j;

         printf("%u\n", ans);
     }
 }

@Author: YouSiki