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  • Codeforces 1105C: Ayoub and Lost Array(递推)

    time limit per test: 1 second
    memory limit per test: 256 megabytes
    input: standard input
    output: standard output

    Ayoub had an array aa of integers of size nn and this array had two interesting properties:

    • All the integers in the array were between ll and rr (inclusive).
    • The sum of all the elements was divisible by 33.

    Unfortunately, Ayoub has lost his array, but he remembers the size of the array nn and the numbers ll and rr, so he asked you to find the number of ways to restore the array.

    Since the answer could be very large, print it modulo 109+710^9+7 (i.e. the remainder when dividing by 109+710^9+7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 00.

    Input

    The first and only line contains three integers nn, ll and rr (1n2105,1lr109)(1≤n≤2⋅10^5,1≤l≤r≤10^9) — the size of the lost array and the range of numbers in the array.

    Output

    Print the remainder when dividing by 109+710^9+7 the number of ways to restore the array.

    Examples

    input
    2 1 3
    output
    3
    input
    3 2 2
    output
    1
    input
    9 9 99
    output
    711426616

    Note

    In the first example, the possible arrays are : [1,2],[2,1],[3,3][1,2],[2,1],[3,3].

    In the second example, the only possible array is [2,2,2][2,2,2].

    题意

    给出三个整数n,l,rn,l,r,要求在[l,r][l,r]之间的数组成的长度为nn的序列的和能够整除33

    AC代码

    #include <stdio.h>
    #include <string.h>
    #include <iostream>
    #include <algorithm>
    #include <math.h>
    #include <limits.h>
    #include <map>
    #include <stack>
    #include <queue>
    #include <vector>
    #include <set>
    #include <string>
    #include <time.h>
    #define ll long long
    #define ull unsigned long long
    #define ms(a,b) memset(a,b,sizeof(a))
    #define pi acos(-1.0)
    #define INF 0x7f7f7f7f
    #define lson o<<1
    #define rson o<<1|1
    #define bug cout<<"-------------"<<endl
    #define debug(...) cerr<<"["<<#__VA_ARGS__":"<<(__VA_ARGS__)<<"]"<<"
    "
    const double E=exp(1);
    const int maxn=1e6+10;
    const int mod=1e9+7;
    using namespace std;
    ll dp[3][maxn];
    int n, l, r, t ;
    int main(int argc, char const *argv[])
    {
    	ios::sync_with_stdio(false);
    	cin>>n>>l>>r;
    	int num0,num1,num2;//余数为i的个数
    	num0=r/3-(l-1)/3;
    	num1=r/3+(r%3>=2)-(l-1)/3-((l-1)%3>=2);
    	num2=r-l+1-num0-num1;
    	dp[0][1]=num0;
    	dp[1][1]=num1;
    	dp[2][1]=num2;
    	for(int i=2;i<=n;i++)
    	{
    		dp[0][i]=((dp[0][i-1]%mod*(num0%mod))%mod+(dp[1][i-1]%mod*num2%mod)%mod+(dp[2][i-1]%mod*num1%mod)%mod)%mod;
    		dp[1][i]=((dp[0][i-1]%mod*(num1%mod))%mod+(dp[1][i-1]%mod*num0%mod)%mod+(dp[2][i-1]%mod*num2%mod)%mod)%mod;
    		dp[2][i]=((dp[0][i-1]%mod*(num2%mod))%mod+(dp[1][i-1]%mod*num1%mod)%mod+(dp[2][i-1]%mod*num0%mod)%mod)%mod;
    	}
    	cout<<dp[0][n]%mod<<endl;
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Friends-A/p/10324307.html
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