Moamen and Ezzat are playing a game. They create an array a of n non-negative integers where every element is less than 2k.
Moamen wins if a1&a2&a3&…&an≥a1⊕a2⊕a3⊕…⊕an.
Here & denotes the bitwise AND operation, and ⊕ denotes the bitwise XOR operation.
Please calculate the number of winning for Moamen arrays a.
As the result may be very large, print the value modulo 1000000007 (109+7).
Input
The first line contains a single integer t (1≤t≤5)— the number of test cases.
Each test case consists of one line containing two integers n and k (1≤n≤2⋅105, 0≤k≤2⋅105).
Output
For each test case, print a single value — the number of different arrays that Moamen wins with.
Print the result modulo 1000000007 (109+7).
Example
input
3 3 1 2 1 4 0
output
5 2 1
Note
In the first example, n=3, k=1. As a result, all the possible arrays are [0,0,0], [0,0,1], [0,1,0], [1,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].
Moamen wins in only 5 of them: [0,0,0], [1,1,0], [0,1,1], [1,0,1], and [1,1,1].
题意:
用 < 2k的数填充到长度为n的数组中,要使得数组中所有数& >= ^,问方案数
显然,当k == 1的时候,答案为1,只有当所有的数全为1的情况才可以满足题意
对于其他的情况{
用小于2k的数进行填充,那么说明填充的数字的二进制位数最多可以有k kk位,
从数的个数的角度来说,奇数个1^ 之后的结果是1,偶数个1^ 之后的结果是0,而对于0来讲,不管多少个0,^起来结果都是0
只要有一个0,那么说这一位上&之后就是0,而为了让^为0,那么只能够选择偶数个1
如果某一位上&为1,那么^为1的情况需要讨论n的奇偶性
{
当n为奇数,全1时&的结果是1,^的结果也是1
当n为偶数,全1时&的结果为1,^运算结果为0,这时显然这一位已经大于异或了,由于后面的权值小,所以可以任意选择
}
}
Code:
ll fact[maxn], infact[maxn]; void init() { fact[0] = infact[0] = 1; for (int i = 1; i <= 200000; i++) { fact[i] = fact[i - 1] * i % mod; infact[i] = infact[i - 1] * qPow(i, mod - 2) % mod; } } ll C(ll n, ll m) { return fact[n] * infact[n - m] % mod * infact[m] % mod; } ll n, m; int main() { init(); int T = read; while (T--) { ll ans = 0; n = read, m = read; if (m == 0) { puts("1"); continue; } if (n % 2) { ans = qPow((qPow(2, n - 1) + 1) % mod, m); /** ll t = 1; for (int i = 0; i <= n - 1; i += 2) { t = (t + C(n,i)) % mod; } ans = qPow(t,m); **/ } else { ll t = 0; /** for (int i = 0; i <= n - 2; i += 2) { t = (t + C(n, i)) % mod; } ans = qPow(t, m); **/ t = qPow(2, n - 1) - 1; ans = qPow(t, m); for (int i = 1; i <= m; i++) { ans = (ans + qPow(t, i - 1) * qPow(2, n * (m - i))) % mod; } } printf("%lld\n", ans); } return 0; }
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