Codeforces Round #533 (Div. 2) C. Ayoub and Lost Array 【dp】
传送门:http://codeforces.com/contest/1105/problem/C
C. Ayoub and Lost Array
1 second
256 megabytes
standard input
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+7109+7 (i.e. the remainder when dividing by 109+7109+7). In case there are no satisfying arrays (Ayoub has a wrong memory), print 00.
The first and only line contains three integers nn, ll and rr (1≤n≤2⋅105,1≤l≤r≤1091≤n≤2⋅105,1≤l≤r≤109) — the size of the lost array and the range of numbers in the array.
Print the remainder when dividing by 109+7109+7 the number of ways to restore the array.
2 1 3
3
3 2 2
1
9 9 99
711426616
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].
题意概括:
要求构造一个长度为 N 的序列,
要求:
1、序列里的数由 【L, R】区间里的数构成。
2、序列里的数值和要能整除 3
解题思路:
一开始还傻傻地以为有什么神奇的规律.....
其实是一道 DP
状态: dp[ i ][ k ] 累积到当前序列第 i 位的数值和 余 k 的方案数
因为要能整除 3 ,所以 k 只能取 0, 1, 2;
sumi 为 区间 【L,R】的模 3 == i 的值的数量
转移方程:
dp[ i ][ 0 ] = dp[i-1][0]*sum0 + dp[i-1][1]*sum2 + dp[i-1][2]*sum1;
dp[ i ][ 1 ] = dp[i-1][0]*sum1 + dp[i-1][1]*sum0 + dp[i-1][2]*sum2;
dp[ i ][ 2 ] = dp[i-1][0]*sum2 + dp[i-1][1]*sum1 + dp[i-1][2]*sum0;
AC code:
#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const LL MOD = 1e9+;
const int MAXN = 2e5+;
LL ans;
LL dp[MAXN][]; int main()
{
LL N, L, R;
LL it0 = , it1 = , it2 = ;
scanf("%I64d %I64d %I64d", &N, &L, &R);
LL len = R-L+;
LL c = len/3LL, d =len%3LL;
it0 = c; it1 = c; it2 = c;
if(d){
LL t = d==?:;
if(L%==) it0++, it1+=t;
else if(L% == ) it1++, it2+=t;
else it2++,it0+=t;
} dp[][] = it0;
dp[][] = it1;
dp[][] = it2; for(int i = ; i <= N; i++){
dp[i][] = ((dp[i-][]*it0)%MOD + (dp[i-][]*it2)%MOD + (dp[i-][]*it1)%MOD)%MOD; dp[i][] = ((dp[i-][]*it0)%MOD + (dp[i-][]*it1)%MOD + (dp[i-][]*it2)%MOD)%MOD; dp[i][] = ((dp[i-][]*it0)%MOD + (dp[i-][]*it1)%MOD + (dp[i-][]*it2)%MOD)%MOD; } printf("%I64d\n", dp[N][]%MOD);
return ; }
Codeforces Round #533 (Div. 2) C. Ayoub and Lost Array 【dp】的更多相关文章
- Codeforces Round #533(Div. 2) C.Ayoub and Lost Array
链接:https://codeforces.com/contest/1105/problem/C 题意: 给n,l,r. 一个n长的数组每个位置可以填区间l-r的值. 有多少种填法,使得数组每个位置相 ...
- Codeforces Round #533 (Div. 2) C. Ayoub and Lost Array(递推)
题意: 长为 n,由 l ~ r 中的数组成,其和模 3 为 0 的数组数目. 思路: dp[ i ][ j ] 为长为 i,模 3 为 j 的数组数目. #include <bits/stdc ...
- Codeforces Round #540 (Div. 3) F1. Tree Cutting (Easy Version) 【DFS】
任意门:http://codeforces.com/contest/1118/problem/F1 F1. Tree Cutting (Easy Version) time limit per tes ...
- Codeforces Round #680 (Div. 2, based on Moscow Team Olympiad)【ABCD】
比赛链接:https://codeforces.com/contest/1445 A. Array Rearrangment 题意 给定两个大小均为 \(n\) 的升序数组 \(a\) 和 \(b\) ...
- Codeforces Round #555 (Div. 3) C2. Increasing Subsequence (hard version)【模拟】
一 题面 C2. Increasing Subsequence (hard version) 二 分析 需要思考清楚再写的一个题目,不能一看题目就上手,容易写错. 分以下几种情况: 1 左右两端数都小 ...
- Codeforces Round #561 (Div. 2) A Tale of Two Lands 【二分】
A Tale of Two Lands 题目链接(点击) The legend of the foundation of Vectorland talks of two integers xx and ...
- Codeforces Round #533 (Div. 2)题解
link orz olinr AK Codeforces Round #533 (Div. 2) 中文水平和英文水平都太渣..翻译不准确见谅 T1.给定n<=1000个整数,你需要钦定一个值t, ...
- Codeforces Round #533 (Div. 2) C.思维dp D. 多源BFS
题目链接:https://codeforces.com/contest/1105 C. Ayoub and Lost Array 题目大意:一个长度为n的数组,数组的元素都在[L,R]之间,并且数组全 ...
- Codeforces Round #533 (Div. 2) Solution
A. Salem and Sticks 签. #include <bits/stdc++.h> using namespace std; #define N 1010 int n, a[N ...
随机推荐
- 使用Charles为Android设备抓取https请求的包
之前开发的Android APP使用的都是http请求,之后改成了https,就出现了以下情况,无法正常读取抓取的内容 找了好多资料说法大概差不多,照着弄,结果出现如下情况,后来发现这种情况其实是手机 ...
- .net HttpClient的使用
在程序用调用 Http 接口.请求 http 资源.编写 http 爬虫等的时候都需要在程序集中进行 Http 请 求. 很多人习惯的 WebClient.HttpWebRequest 在 TPL ...
- 异步http请求的实现
这是我自己在某论坛上发的一篇水贴:http://www.sufeinet.com/thread-9275-1-2.html,原理和解释,我就直接重发一遍在自己博客上了. 时隔一个月 回来把之前的坑填 ...
- Rabbit简单队列模式
1 <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/ ...
- 13. Roman to Integer 罗马数字转化为阿拉伯数字(indexOf ()和 toCharArray())easy
Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M. Symbol Value I 1 ...
- 中南oj 1216: 异或最大值 数据结构
1216: 异或最大值 Time Limit: 2 Sec Memory Limit: 128 MB Submit: 98 Solved: 29 [Submit][Status][Web Boar ...
- 反汇编调试Android
https://code.google.com/p/android/issues/detail?id=73076 http://my.unix-center.net/~Simon_fu/?p=527 ...
- java 从Excel 输出和输入
本文实现了使用java 从数据库中获得对象,并存入集合中, 然后输出到Excel,并设置样式 package com.webwork; import java.io.File; import java ...
- csharp: Converting chinese character to Unicode
Function chinese2unicode(Str) Dim Str_one:Str_one = "" Dim Str_unicode:Str_unicode = " ...
- python 中 \n 和转义r的作用和\r的实际应用
我们先看看这张转义字符图: 1. 知识储备 \r 表示将光标的位置回退到本行的开头位置 \b 表示将光标的位置回退一位 在 python 语言中, 使用 print 打印输出时,默认是会进行换行的.如 ...