Ayoub and Lost Array

https://codeforces.com/contest/1105/problem/C

C++版本一

题解：DP+矩阵快速幂

参考文章：https://blog.csdn.net/weixin_43272781/article/details/85939539

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,l,r,q;
int ans,cnt,flag,temp;
int a[N];
char str;

int tmp[N][N];
void multi(ll a[][3],ll b[][3])
{
    memset(tmp,0,sizeof tmp);
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
            for(int k=0;k<3;k++)
                tmp[i][j]=(tmp[i][j]+a[i][k]*b[k][j])%MOD;
    for(int i=0;i<3;i++)
        for(int j=0;j<3;j++)
            a[i][j]=tmp[i][j];
}
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    scanf("%d%d%d",&n,&l,&r);
    a[0]=a[1]=a[2]=r/3;
    if(r%3==1)
        a[1]++;
    if(r%3==2)
        a[1]++,a[2]++;
    a[0]-=(l-1)/3;
    a[1]-=(l-1)/3;
    a[2]-=(l-1)/3;
    if((l-1)%3==1)
        a[1]--;
    if((l-1)%3==2)
        a[1]--,a[2]--;
//    b[0]=b[0]*a[0]+b[1]*a[2]+b[2]*a[1];
//    b[1]=b[0]*a[1]+b[1]*a[0]+b[2]*a[2];
//    b[2]=b[0]*a[2]+b[1]*a[1]+b[2]*a[0];
    ll b[5][3]={a[0],a[1],a[2],a[2],a[0],a[1],a[1],a[2],a[0]};
    ll c[5][3]={a[0],a[1],a[2],a[2],a[0],a[1],a[1],a[2],a[0]};
    n--;
    while(n){
        if(n&1){
            multi(b,c);
        }
        multi(c,c);
        n/=2;
    }
    cout<<b[0][0]<<endl;
    //cout << "Hello world!" << endl;
    return 0;
}

C++版本二

题解：数位DP

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<algorithm>
#define ll long long
#define pb push_back
#define test printf("here!!!")
using namespace std;
const int mx=2e5+10;
const ll mod=1e9+7;
ll a,b,c;
ll f[mx][3];
int main()
{
    int n,l,r;
    scanf("%d%d%d",&n,&l,&r);
    int ls=l;
    int rs=r;
    int ys;
    while (ls%3!=0)
    {
        ls++;
    }
    while (rs%3!=0)
    {
        rs--;
    }
    if (r-l<3)
    {
        for (int i=l;i<=r;++i)
        {
            ys=i%3;
            if (ys==1) b++;
            else if (ys==2) c++;
            else if (ys==0) a++;
        }
    }
    else
    {
        if (ls<=rs&&ls<=r&&rs>=l)
        {
            a=rs/3-ls/3+1;
            b=a-1;
            c=a-1;
        }
        for (int i=l;i<ls;++i)
        {
            ys=i%3;
            if (ys==1) b++;
            else if (ys==2) c++;
            else if (ys==0) a++;
        }
        for (int i=rs+1;i<=r;++i)
        {
            ys=i%3;
            if (ys==1) b++;
            else if (ys==2) c++;
            else if (ys==0) a++;
        }
    }
    f[1][0]=a;
    f[1][1]=b;
    f[1][2]=c;
    for (int i=2;i<=n;++i)
    {
        f[i][0]=(f[i-1][1]*c%mod+f[i-1][2]*b%mod+f[i-1][0]*a%mod)%mod;
        f[i][1]=(f[i-1][1]*a%mod+f[i-1][2]*c%mod+f[i-1][0]*b%mod)%mod;
        f[i][2]=(f[i-1][1]*b%mod+f[i-1][2]*a%mod+f[i-1][0]*c%mod)%mod;
    }
    printf("%I64d\n",f[n][0]);
}