HDU 5363 elements for a set of 1~n how many subsets of elements and for even-thinking-(Fast power modulo)

Test instructions: A set has element 1~n that satisfies the condition that all elements of a subset are of an even number, asking how many such subsets

Analysis:

A permutation combination problem. Elements and is even, then the odd number must be adjusted even numbers, even if it does not matter, so even there is 2^ (N/2) Seed selection method, and then multiplied by odd (C ((n+1)/2,0) +c ((n+1)/2,2) ... Seed selection method, minus one, remove the empty set, note that the above is used when the odd number (n+1)/2 (here is the division of the downward rounding), is a combination of n is even and N is an odd two cases.

Combination Number properties: C (n,1) +c (n,3) +....=c (n,2) +c (n,4) +....=1/2* (c (n,1) +c (n,2) + ... C (N.N)) =1/2* (2^n) =2^ (n-1) (n is an odd even number that satisfies this property)

Now Bashi the sub-merge (where the division is true division, so N is an odd time to be +1 or 1 to divide):

When 1.N is even, even there are 2^ (N/2), Odd 2^ (N/2)/2 kinds of extraction, integrated ans=2^ (n-1)-1

When 2.N is odd, even there are 2^ ((n-1)/2), Odd 2^ ((n+1)/2), Integrated ans=2^ (N-1) 1

The rest is the quick power to take the mold, this method is very basic to remember it.

Code:

#include <cstdio> #include <iostream> #define INF 1000000007using namespace Std;int t;long long N;long long ans ; Long long f (long long a,long long B,long long c) {    long long t = 1;    while (b) {        if (b&1) t= (t*a)%inf;        b>>=1;        A= (a*a)%inf;    }    return t; }int Main () {    scanf ("%d", &t);    while (t--) {        scanf ("%i64d", &n);            Ans=f (2,n-1,inf);            Ans-=1;            printf ("%i64d\n", ans);}    }

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HDU 5363 elements for a set of 1~n how many subsets of elements and for even-thinking-(Fast power modulo)