B.Icebound and Sequence

链接:https://ac.nowcoder.com/acm/contest/903/B

题意:

Icebound hates math. But Imp loves math. One day, Imp gave icebound a problem.

The problem is as follows.

S=(ni=1qi mod pS=(∑i=1nqi) mod p

For given q,n,p, you need to help icebound to calculate the value of S.

思路:

等比数列求和。

因为考虑到取余,所以不能直接算。

令S(n) 为等比数列前n项和。

若n为偶数:

  S(n) = S(n/2) + S(n/2)*a^(n/2) (因为第i(i <= n/2)项和i+n/2项存在第i项乘a^(n/2)等以第i+n/2项的值。

若n为奇数:

  S(n) = S(n/2) + S(n/2)*a^(n/2) + a^n

代码:

#include <bits/stdc++.h>
using namespace std;

typedef long long LL;

LL n, q, p;

LL QM(LL a, LL b, LL m)
{
    LL res = 1;
    while (b)
    {
        if (b&1)
            res = (res*a)%m;
        a = (a*a)%m;
        b >>= 1;
    }
    return res;
}

LL GetR(int t)
{
    if (t == 1)
        return n%p;
    if (t%2 == 0)
        return (GetR(t/2)+(GetR(t/2)*QM(n, t/2, p))%p)%p;
    else
        return ((GetR(t/2)+(GetR(t/2)*QM(n, t/2, p))%p)%p+QM(n, t, p))%p;
}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        cin >> n >> q >> p;
        cout << GetR(q) << endl;
    }

    return 0;
}

  

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转载自www.cnblogs.com/YDDDD/p/10923486.html
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