A^X mod PIt's easy for ACMer to calculate A^X mod P. Now given seven integers n, A, K, a, b, m, P, and a function f(x) which defined as following.f(x) = K, x = 1f(x) = (a*f(x-1) + b)%m , x > 1Now, Your task is to calculate( A^(f(1)) + A^(f(2)) + A^(f(3)) + ...... + A^(f(n)) ) modular P.输入In the first line there is an integer T (1 < T <= 40), which indicates the number of test cases, and then T test cases follow. A test case contains seven integers n, A, K, a, b, m, P in one line.1 <= n <= 10^60 <= A, K, a, b <= 10^91 <= m, P <= 10^9输出For each case, the output format is “Case #c: ans”.c is the case number start from 1.ans is the answer of this problem.
时间: 2024-04-17 21:26:12 浏览: 171
您好!根据您的描述,您需要计算表达式( A^(f(1)) + A^(f(2)) + A^(f(3)) + ...... + A^(f(n)) ) mod P的结果。根据给定的参数n, A, K, a, b, m, P以及函数f(x)的定义,我将为您提供一个解决方案。
首先,我们需要计算函数f(x)的值。根据定义,当x=1时,f(x)等于K;当x>1时,f(x)等于(a*f(x-1) + b)%m。
然后,我们可以使用循环来计算表达式的结果。从x=1到x=n,我们将每个f(x)对应的A^(f(x)) mod P加到结果中。最后,我们将结果对P取模,并输出答案。
下面是一个示例代码实现:
```python
def calculate_expression(n, A, K, a, b, m, P):
# 计算函数f(x)的值
f = [0] * (n+1)
f[1] = K
for x in range(2, n+1):
f[x] = (a*f[x-1] + b) % m
# 计算表达式的结果
result = 0
for x in range(1, n+1):
result += pow(A, f[x], P)
# 对结果取模并输出答案
ans = result % P
return ans
# 处理多个测试用例
T = int(input())
for case in range(1, T+1):
n, A, K, a, b, m, P = map(int, input().split())
ans = calculate_expression(n, A, K, a, b, m, P)
print("Case #{}: {}".format(case, ans))
```
希望这个解决方案能够帮助到您!如果您有任何其他问题,请随时提问。
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