UVA12653 Buses-白红宇

UVA12653 Buses

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发布时间：2019-06-22

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Problem H

Buses

File: buses.[c|cpp|java]

Programming competitions usually require infrastructure and organization on the part of those

responsible. A problem that frequently must be solved is regarding transportation. While participating

in a recent competition, Ricardinho watched the buses and micro-buses used in the transportation of

competitors, all lined up one behind the other as competitors disembarked. The vehicles were all from

the same company, although had different paintings. Ricardinho began to wonder how many ways

that line could be formed using buses and minibuse from that company.

Each bus is 10 meters long, each minibus is 5 meters long. Given the total length of a line of buses

and minibuses, and the number of different colors each buse or minibus may be painted, Ricardinho

wants to know in how many ways such a line can be formed.

Input

The input contains several test cases. Each test case is composed of a single line, containing three

integers N, K and L, representing respectively the total length, in meters, of the line Ricky is con-sidering, K indicates the number of different colors for micro-buses, and L represents the number of

different colors for buses. Note that, as integers N , K and L may be very large, the use of 64 bits

integers is recommended.

Output

As the number of different ways of forming the line can be very large, Ricardinho is interested in the

last 6 digits of that quantity. Thus, your for each test case your program must produce a single line

containing exactly 6 digits, corresponding to the last digits of the solution.

Restrictions

• 5 ≤ N ≤ 10

and N is multiple of 5

• 1 ≤ K ≤ 10

• 1 ≤ L ≤ 10

Examples

Input

25 5 5

5 1000 1000

20 17 31

15 9 2

Output

006000

001000

111359

000765

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                 #define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))using namespace std ;typedef long long LL ;const int M=200008 ;const LL Mod=1000000 ;LL A[M] ;struct Mat{ LL num[3][3] ; Mat(){} ; Mat(int a11 ,int a12 ,int a21 ,int a22){ num[1][1]=a11 ; num[1][2]=a12 ; num[2][1]=a21 ; num[2][2]=a22 ; } Mat operator *(Mat &B){ Mat ans(0,0,0,0) ; for(int i=1 ;i<=2 ;i++) for(int j=1;j<=2 ;j++) for(int k=1;k<=2; k++){ ans.num[i][j]=ans.num[i][j]+num[i][k]*B.num[k][j] ; if(ans.num[i][j]>=Mod) ans.num[i][j]%=Mod ; } return ans ; }};Mat Pow(Mat X ,LL y){ Mat ans=Mat(1,0,0,1) ; for(;y;y>>=1){ if(y&1) ans=ans*X ; X=X*X ; } return ans ;}LL a[3] ;int main(){ LL N ,K ,L ; while(cin>>N>>K>>L){ N/=5 ; Mat A(K%Mod ,L%Mod ,1 ,0) ; a[1]=K%Mod ; a[2]=(((K%Mod)*(K%Mod))%Mod+L%Mod)%Mod ; if(N==1) printf("%06d\n",(int)a[1]) ; else if(N==2) printf("%06d\n",(int)a[2]) ; else{ A=Pow(A,N-2) ; LL ans=(A.num[1][1]*a[2]%Mod+A.num[1][2]*a[1]%Mod)%Mod ; printf("%06d\n",(int)ans) ; } } return 0 ;}