（矩阵快速幂）hdu4686 Arc of Dream_arc of dream 矩阵快速幂-优快云博客

本文链接：https://blog.youkuaiyun.com/ziqi898004/article/details/77073943

（写了三天博客之后，终于学会了插入链接~hhhhhh）题目链接如下：

传送门——hdu4686 Arc of Dream

（也学会了复制题目！23333）题目描述：（居然是watashi大佬出的题，赶紧膜一发....）

Problem Description

An Arc of Dream is a curve defined by following function:

where
a ₀ = A0
a _i = a _i-1*AX+AY
b ₀ = B0
b _i = b _i-1*BX+BY
What is the value of AoD(N) modulo 1,000,000,007?

Input

There are multiple test cases. Process to the End of File.
Each test case contains 7 nonnegative integers as follows:
N
A0 AX AY
B0 BX BY
N is no more than 10 ¹⁸, and all the other integers are no more than 2×10 ⁹.

Output

For each test case, output AoD(N) modulo 1,000,000,007.

Sample Input

Sample Output

Author

Zejun Wu (watashi)

题目很裸，就是求一个函数F（n）第n项的值，思路很简单，由题中所给关系求出递推式：aibi = (AX*BX)*(ai-1*bi-1) + (AX*BY)*ai-1 + (AY*BX)*bi-1 + (AY*BY)，并且Fi = Fi-1 + ai*bi，因此相当于Fn = (AX*BX)*(ai-1*bi-1) + (AX*BY)*ai-1 + (AY*BX)*bi-1 + (AY*BY)，用矩阵相乘简化这个递推式，设计一个五维向量（ai*bi, ai, bi, 1, Fi），他可以由它的前一项与一个常数矩阵相乘得出，再利用矩阵快速幂，注意数据范围是long long，顺利ac~

P.S：注意要加一句，n=0时输出结果0，并且，千万不要自作聪明地认为这句判断加在哪都无所谓！！当时不注意，写了这样一个极难发现的bug，WA了6发，WA到一度怀疑人生相信玄学，结果发现是n=0输出的时间出了问题T_T mdzz

代码如下~（写的很恶心但改成这样就ac了再改又wa 于是不敢动了orz）

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

const ll mod = 1000000007;

ll n, a0, ax, ay, b0, bx, by;

struct mat {
    ll a[5][5];
};

void print(mat ans) {
    for(int i = 0;i < 5;i++) {
        for(int j = 0;j < 5;j++)
            printf("%d ", ans.a[i][j]);
        printf("\n");
    }
    return;
}

mat init0() {//初始化
    mat tt;
    for(int i = 0;i < 5;i++)
        for(int j = 0;j < 5;j++) tt.a[i][j] = 0;
    return tt;
}

mat init1() {//初始化
    mat tt;
    for(int i = 0;i < 5;i++) {
        for(int j = 0;j < 5;j++) {
            if(i == j)
                tt.a[i][j] = 1;
            else
                tt.a[i][j] = 0;
        }
    }
    return tt;
}

mat mul(mat x, mat y) {
    mat tmp;
    for(int i = 0;i < 5;i++) {
        for(int j = 0;j < 5;j++) {
            tmp.a[i][j] = 0;
            for(int k = 0;k < 5;k++) {
                tmp.a[i][j] = (tmp.a[i][j] + x.a[i][k]*y.a[k][j]%mod) % mod;
            }
        }
    }
    return tmp;
}

mat qpow(mat tt, ll b) {
    mat res = init1();
    while(b) {
        if(b & 1) {
            res = mul(res, tt);
        }
        b >>= 1;
        tt = mul(tt, tt);
    }
    return res;
}

int main() {
    while(~scanf("%I64d", &n)) {
        scanf("%I64d %I64d %I64d", &a0, &ax, &ay);
        scanf("%I64d %I64d %I64d", &b0, &bx, &by);
        if(n == 0)
            printf("0\n");
        else {
            mat coef = init0();
            coef.a[0][0] = ( (ax%mod) * (bx%mod) ) % mod;
            coef.a[0][1] = ( (ax%mod) * (by%mod) ) % mod;
            coef.a[0][2] = ( (ay%mod) * (bx%mod) ) % mod;
            coef.a[0][3] = ( (ay%mod) * (by%mod) ) % mod;
            coef.a[1][1] = ax % mod;
            coef.a[1][3] = ay % mod;
            coef.a[2][2] = bx % mod;
            coef.a[2][3] = by % mod;
            coef.a[3][3] = 1;
            coef.a[4][0] = ( (ax%mod) * (bx%mod) ) % mod;
            coef.a[4][1] = ( (ax%mod) * (by%mod) ) % mod;
            coef.a[4][2] = ( (ay%mod) * (bx%mod) ) % mod;
            coef.a[4][3] = ( (ay%mod) * (by%mod) ) % mod;
            coef.a[4][4] = 1;
            coef = qpow(coef, n-1);
            mat mat1 = {((a0%mod)*(b0%mod))%mod, 0, 0, 0, 0,
                        a0%mod,    0, 0, 0, 0,
                        b0%mod,    0, 0, 0, 0,
                        1,     0, 0, 0, 0,
                        ((a0%mod)*(b0%mod))%mod, 0, 0, 0, 0
                        };
            mat ans = mul(coef, mat1);
            printf("%I64d\n", ans.a[4][0] % mod);
        }
    }
    return 0;
}