1353 the game of the Great Gods

Source: Internet
Author: User

1353 The game of the Great Gods

time limit: 1 sspace limit: 128000 KBtitle level: Golden Gold SolvingTitle Description Description

In that remote room, there is a magic lattice. For the sake of convenience, we are numbered 1~n. Legend has it that if you put some cards, you can achieve your wishes. The cards have a total of M colors, but the adjacent compartments cannot be placed in the same color card. By not repeating all the combinations, you can summon the Divine Burger @ To solve your great God and fulfill your dream. The SHC classmate who had turned out the records from the ancient books, he set up his card on the night. Now he wants to know how many different combinations are illegal in order to figure out the day the wish was fulfilled. But our SHC classmates are busy posing cards, and this task will be handed to you naturally.

Enter a description Input Description

Enter a two integer m,n.

1<=m<=10^8,1<=n<=10^12

Output description Output Description

Number of combinations, modulus 10086 take-up

Sample input Sample Input
2 3
Sample output Sample Output

6

Data range and Tips Data Size & Hint

6 states of (000) (001) (011) (100) (110) (111)

60% of data n<=100 0000

God Burger is a famous sister, not black her XD

Category labels Tags Click here to expandArithmetical
//Answer: (m^n-m* (m-1) ^ (n-1))%10086#include <cstdio>#include<iostream>using namespacestd;#defineMoD 10086Long LongM,n,p1,p2,ans;Long Longf[1001];/*Long Long Quick_pow (long long X,long long n) {if (n==0) return 1; else{while (!) (            n&1)) {n>>=1;        X*=x;    }} long long result=x;    n>>=1;        while (n) {x*=x;        if ((n&1)) {result*=x;    } n>>=1; } return result;*/Long LongQuick_pow (Long LongBLong LongA) {//Note: The fast power inside to take the remainder     if(!a)return 1;//using the idea of dichotomy     if(a&1)return(Quick_pow (b,a-1)%mod*b)%mod;//Odd     Long LongT=quick_pow (b,a/2)%mod;//even     return(t*t)%MoD;}intMain () {CIN>>m>>N; P1=quick_pow (m%mod,n)%mod;//Note that the power cannot be taken, because there is no such formulaP2=quick_pow (M-1)%mod, (n1)) *m%MoD; Ans= (p1-p2+mod*Ten)%MoD; cout<<ans<<Endl; return 0;}

1353 the game of the Great Gods

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