CodeChef-----February Challenge 2018---Broken Clock(极坐标+三角函数递推+矩阵快速幂)-优快云博客

本文针对CodeChef的BrokenClock问题进行了解析，介绍了如何利用矩阵快速幂的方法求解钟表上分针位置的问题，并给出了具体的实现代码。

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链接: https://www.codechef.com/FEB18/problems/BROCLK

Broken Clock Problem Code: BROCLK

Chef has a clock, but it got broken today — the minute hand on Chef's clock doesn't rotate by the angle 2π/3600 each second, but by a different fixed angle x. The coordinates of the center of the clock are (0, 0). The length of the minute hand is l.

One endpoint of the minute hand is always located at the clock center; the other endpoint is initially located at the point (0, l). One second later, Chef observes that this endpoint is at distance d above the x-axis, i.e. the y-coordinate of this endpoint is equal to d.

Chef is curious about where the minute hand will be (specifically, its y-coordinate) after tseconds. Because t can be very large, Chef can't wait for that moment. Please help him!

Input

The first line of the input contains a single integer T denoting the number of test cases. The description of T test cases follows.
The first and only line of each test case contains three space-separated integers l, dand t.

Output

We can prove that for the given constraints, the y-coordinate of the end of the minute hand can always be written as a rational number p / q, where gcd(p, q) = gcd(q, 10⁹ + 7) = 1. Let's denote the modular inverse of q (it's guaranteed that the modular inverse exists and is unique) by r.

For each test case, print a single line containing one number (p · r) modulo 10⁹ + 7.

Constraints

1 ≤ T ≤ 10⁵
1 ≤ d < l ≤ 10⁹
1 ≤ t ≤ 10¹⁸

Subtasks

Subtask #1 (5 points): t ≤ 3

Subtask #2 (15 points): t is a power of 2, i.e. t = 2^p for some p ≥ 0

Subtask #3 (40 points): sum of t over all test cases ≤ 10⁶

Subtask #4 (40 points): original constraints

Example

Input:

3
4 2 1
4 2 2
4 2 3

Output:

2
1000000005
1000000003

贴个代码

吃个教训，在february challenge 期间就传了代码，然后有人直接就这个代码交了，被codechef发邮件通知掉rating了，以后不敢了

#include <bits/stdc++.h>
#define mst(a,b)    memset((a),(b), sizeof a)
#define lowbit(a)   ((a)&(-a))
#define IOS         ios::sync_with_stdio(0);cin.tie(0);
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn=1e6+10;
const ll INF = 1LL<<60;
const int N=2;
ll qpow(ll a,ll b){
    ll ret=1;
    while(b){
        if(b&1)ret=ret*a%mod;
        b>>=1;a=a*a%mod;
    }
    return ret;
}
struct matrix{
    ll mat[N][N];
    matrix operator*(const matrix&m)const{
        matrix tmp;
        for(int i=0;i<N;++i){
            for(int j=0;j<N;++j){
                tmp.mat[i][j]=0;
                for(int k=0;k<N;++k){
                    tmp.mat[i][j]+=mat[i][k]*m.mat[k][j]%mod;
                    tmp.mat[i][j]%=mod;
                }
            }
        }
        return tmp;
    }
};
void solve(ll d,ll l,ll t,ll&a,ll&b){
    ll gg=__gcd(d,l);d/=gg,l/=gg;l%=mod,d%=mod;

    b=qpow(l,t);--t;
    matrix m,ans;
    mst(m.mat,0);mst(ans.mat,0);
    for(int i=0;i<N;++i)ans.mat[i][i]=1;
    m.mat[0][0]=2*d%mod;m.mat[0][1]=-(l*l%mod);
    m.mat[1][0]=1;
    while(t){
        if(t&1)ans=ans*m;
        t>>=1;
        m=m*m;
    }
    a=(ans.mat[0][0]*d%mod+ans.mat[0][1]%mod)%mod;
    a=(a+mod)%mod;
}
int main(){
#ifdef local
freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
#endif
    int t;scanf("%d",&t);
    while(t--){
        ll d,l,t;scanf("%lld%lld%lld",&l,&d,&t);
        ll a,b;solve(d,l,t,a,b);
//        cout<<a<<" "<<b<<endl;
        printf("%lld\n",l*a%mod*qpow(b,mod-2)%mod);
    }
    return 0;
}