Description
It'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 = 1
f(x) = (a*f(x-1) + b)%m , x > 1
Now, Your task is to calculate
( A^(f(1)) + A^(f(2)) + A^(f(3)) + ...... + A^(f(n)) ) modular P.
Input
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 <= 10^6
0 <= A, K, a, b <= 10^9
1 <= m, P <= 10^9
Output
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.
Sample
Input
2
3 2 1 1 1 100 100
3 15 123 2 3 1000 107
Output
Case #1: 14
Case #2: 63
Hint
Source
2013年山东省第四届ACM大学生程序设计竞赛
附上一段会T的代码：
#include <stdio.h>
#include <vector>
#include <
string.h>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
#define ll long long
const int inf =
0xffffff;
const int maxn = 1e6+
8;
int n, t;
ll A, K, a, b, m, P, f[maxn], fi[maxn];
ll qsm(ll x, ll y, ll z)
{
ll ans =
1;
while(y)
{
if(y&
1)
{
ans = ans*x%
z;
}
x = x*x%
z;
y >>=
1;
}
return ans;
}
int main()
{
scanf("%d", &
t);
int ga =
t;
while(t--
)
{
ll sum;
scanf("%d%lld%lld%lld%lld%lld%lld", &n, &A, &K, &a, &b, &m, &
P);
ll miao = K;
//存下第一项，f(x) = k
sum = qsm(A, K, P);
//存下第一项 A^(f(1))%P
for(
int i =
1; i<n; i++
)
{
miao = (a*miao+b)%m;
//由题可知，后一项等于 a*前一项+b，这里是依次算出以后的每一个f(i)
sum = (sum+qsm(A, miao, P))%P;
//这里是依次算出A^(f(i))%P
}
printf("Case #%d: %lld\n", ga-
t, sum);
}
return 0;
}
转载于:https://www.cnblogs.com/RootVount/p/10679767.html
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