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  • 位运算 2013年山东省赛 F Alice and Bob

    题目传送门

     1 /*
     2     题意: 求(a0*x^(2^0)+1) * (a1 * x^(2^1)+1)*.......*(an-1 * x^(2^(n-1))+1) 式子中,x的p次方的系数
     3     二进制位运算:p = 2 ^ i + 2 ^ j + 2 ^ k  + ...,在二进制表示下就是1的出现
     4            例如:10 的二进制 为1010,10 = 2^3 + 2^1 = 8 + 2,而且每一个二进制数都有相关的a[i],对p移位运算,累计取模就行了
     5 */
     6 #include <cstdio>
     7 #include <cmath>
     8 #include <algorithm>
     9 #include <iostream>
    10 #include <cstring>
    11 using namespace std;
    12 
    13 const int MAXN = 1e6 + 10;
    14 const int INF = 0x3f3f3f3f;
    15 
    16 int main(void)        //F Alice and Bob
    17 {
    18     //freopen ("F.txt", "r", stdin);
    19     
    20     int t;
    21     while (scanf ("%d", &t) == 1)
    22     {
    23         while (t--)
    24         {
    25             int n, a[100], q;
    26             long long p;
    27             scanf ("%d", &n);
    28             for (int i=0; i<n; ++i)
    29             {
    30                 scanf ("%d", &a[i]);
    31             }
    32             scanf ("%d", &q);
    33             
    34             while (q--)
    35             {
    36                 scanf ("%lld", &p);
    37                 int tmp = 1;    int i = 0;
    38                 while (p)
    39                 {
    40                     if (i >= n)
    41                     {
    42                         tmp = 0;    break;
    43                     }
    44                     if (p & 1)        tmp *= a[i];
    45                     ++i;
    46                     p >>= 1;
    47                     tmp %= 2012;
    48                 }
    49                 printf ("%d
    ", tmp);
    50             }            
    51         }
    52     }
    53     
    54     return 0;
    55 }
    编译人生,运行世界!
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  • 原文地址:https://www.cnblogs.com/Running-Time/p/4398620.html
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