Homework of English

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本文介绍了一个基于序列的数学问题背景下的英语作业难题，探讨了如何通过算法高效地解决该问题。主要内容包括输入输出格式说明、样例解释及核心算法实现。

Homework of English

The English teacher give Nobita a very very strange homework.
And Nobita am really doesn't konw why teacher gives him such a homwork.
Perhaps that's because of Nobita's English is very bad.
The homework is like this.
There are a integer sequence a[1],a[2],...,a[N] of length N, and two integer P and M.
And then there are Q queries.
Each query contains two integers L and R.
Nobita need to konw the value (a[L]*(P^0)+a[L+1]*(P^1)+a[L+2]*(P^2)+...+a[R]*(P^(R-L)))%M for each query.

And for any two queries L1,R1 and L2,R2,
if L1<L2 then R1<L2 or R1>=R2.

Input

There are lots of test case.
For each test case.
The first line contains integers N,P,M.
The second line contains integers a[0],b,c,d
And the sequence a will be got through this four numbers, that is a[i]=(a[i-1]*b+c)%d for 1<=i<=N.
The Third line contains a integer Q.
The next Q lines each line contain two integers L and R.

(1<=N<=5000000, 2<=P,M<=10^9, 1<=Q<=10^5, 1<=L<=R<=N, 1<=a[0],b,c,d<=10^9)

Output

For each query output one line.
Contain a integer indicates the answer.

Sample Input

5 3 10
3 4 5 6
2
2 4
3 3

Sample Output

5
3

$f[i]=\sum_{x=1}^ia[x]*p^{x-1}$
$ans=\frac{f[R]-f[L-1]}{p^L}$
然后给定的M不一定与P互素, $P^L$ 关于M的乘法逆元不一定存在

前缀和不行，那就考虑一下后缀和……

$f[i]=a[i]+a[i+1]*p+a[i+2]*p^2+...+a[n]^{p-i}$
$ans=f[L]-f[R+1]*p^{r-l+1}$

#include<iostream>
#include<cstdlib>
#include<cstdio>
#include<string>
#include<vector>
#include<deque>
#include<queue>
#include<algorithm>
#include<set>
#include<map>
#include<stack>
#include<ctime>
#include<string.h>
#include<math.h>
#include<list>

using namespace std;

#define ll long long
#define pii pair<int,int>
const int inf = 1e9 + 7;

const int N = 5e6+5;
ll MOD;

ll f[N];

ll qPow(ll a,ll n){
    ll ans=1;
    ll t=a%MOD;
    while(n){
        if(n&1){
            ans=(ans*t)%MOD;
        }
        n>>=1;
        t=(t*t)%MOD;
    }
    return ans;
}

int main()
{
    //freopen("/home/lu/Documents/r.txt","r",stdin);
    ll n,p;
    while(~scanf("%lld%lld%lld",&n,&p,&MOD)){
        ll a0,b,c,d;
        scanf("%lld%lld%lld%lld",&a0,&b,&c,&d);
        f[0]=a0;
        for(int i=1;i<=n;++i){
            f[i]=(f[i-1]*b+c)%d;
        }
        f[n+1]=0;
        for(int i=n-1;i>=0;--i){
            f[i]=(f[i+1]*p+f[i])%MOD;
        }
        int q;
        scanf("%d",&q);
        while(q--){
            int l,r;
            scanf("%d%d",&l,&r);
            ll ans=((f[l]-f[r+1]*qPow(p,r-l+1))%MOD+MOD)%MOD;
            printf("%lld\n",ans);
        }
    }
    return 0;
}