Southwest min da oj (recursive)

Source: Internet
Author: User

My math can't be that hard to push.Time limit (normal/java): ms/9000 MS Run memory limit: 65536 KByte
Total Submissions: 49 Tested by: 24


No problem, how to do it ~

All right, Baidu goes together.

So this problem turned out, only for happy.

Xiao Ming likes to play chess, one day he idle to bored, black and white pieces into a long observation, found black mixed together good ugly ah.

Immediately thought, black and sunspots are not adjacent, placed into a strip, how many kinds of possible?!


Multiple sets of test data (approx. 1w Group)

First line, total number of pieces N (0<n<=1000000)


Only one row, to ensure that sunspots and sunspots are not adjacent, the number of pieces placed in all cases after the result of 1000000007 modulo.

Sample input


Sample output


Analysis: Set A[i] for the total number of pieces of I, b[i] for the total number of white ending, c[i] for the total number of black chess. Then there is a[i]=b[i]+c[i],b[i]=a[i-1],c[i]=b[i-1];

#include <cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<queue>#include<cstdlib>#include<stack>#include<vector>#include<Set>#include<map>#defineLL Long Long#defineMoD 1000000007#defineINF 0x3f3f3f3f#defineN 1000010using namespacestd;intA[n],b[n],c[n];voidinit () {a[1]=2; b[1]=1; c[1]=1;  for(intI=2; i<=1000000; i++) {B[i]=a[i-1];c[i]=b[i-1]; A[i]=b[i]+C[i]; A[i]%=MoD; }}intMain () {intN;    Init ();  while(SCANF ("%d", &n) >0) {printf ("%d", A[n]); }}
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Southwest min da oj (recursive)

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