HDU 3485 count 101

Count 1, 101

Time Limit: 2000/1000 MS (Java/others) memory limit: 65536/32768 K (Java/Others)
Total submission (s): 1114 accepted submission (s): 568

Problem descriptionyou know Yaoyao is fond of his chains. he has a lot of chains and each chain has n diamonds on it. there are two kinds of diamonds, labeled 0 and 1. we can write down the label of diamonds on a chain. so each chain can be written as a sequence consisting of 0 and 1.
We know that chains are different with each other. And their length is exactly N. And what's more, each chain sequence doesn't contain "101" as a substring.
Cocould you tell how would your chains will Yaoyao have at most?

 

Inputthere will be multiple test cases in a test data. for each test case, there is only one number n (n <10000 ). the end of the input is indicated by a-1, which shoshould not be processed as a case.

 

Outputfor each test case, only one line with a number indicating the total number of chains Yaoyao can have at most of length n. The answer shocould be print after module 9997.

 

Sample Input34-1

 

Sample output712 HintWe can see when the length equals to 4. We can have those chains :,

 

Source2010 ACM-ICPC multi-university training Contest (5) -- host by bjtu

 

Recommendzhengfeng is a bit similar to the digital DP, but it is not set to the digital DP [I] [J] [k] to indicate that the last two digits of the I string are I, j does not include 101. So the DP equation has the following:

DP [I] [0] [0] + = DP [I-1] [0] [1] + dp [I-1] [0] [0]
DP [I] [0] [1] + = DP [I-1] [1] [0] + dp [I-1] [1] [1]
DP [I] [1] [0] + = DP [I-1] [0] [0]
DP [I] [1] [1] + = DP [I-1] [1] [0] + dp [I-1] [1] [1]

Finally, retrieve the total number of ans.

#include<cstdio>#include<iostream>#include<cstring>#include<cmath>#include<stdlib.h>#include<algorithm>using namespace std;const int mod=9997;const int MAXN=10000+5;int dp[MAXN][2][2],n,ans[MAXN];void init(){    memset(dp,0,sizeof(dp));    memset(ans,0,sizeof(ans));    ans[0]=0;ans[1]=2;ans[2]=4;    dp[2][0][0]=dp[2][0][1]=dp[2][1][0]=dp[2][1][1]=1;    dp[1][1][0]=dp[1][0][0]=1;    for(int i=3;i<MAXN;i++)    {        dp[i][0][0]=(dp[i][0][0]+(dp[i-1][0][1]+dp[i-1][0][0])%mod)%mod;        dp[i][0][1]=(dp[i][0][1]+(dp[i-1][1][0]+dp[i-1][1][1])%mod)%mod;        dp[i][1][0]=(dp[i][1][0]+(dp[i-1][0][0])%mod)%mod;        dp[i][1][1]=(dp[i][1][1]+(dp[i-1][1][0]+dp[i-1][1][1])%mod)%mod;        ans[i]=(dp[i][0][0]+dp[i][0][1]+dp[i][1][0]+dp[i][1][1])%mod;    }}int main(){    init();    while(scanf("%d",&n)&&n!=-1)    {        printf("%d\n",ans[n]);    }    return 0;}

View code

 

HDU 3485 count 101