HDU 5363 element is 1 ~ Number of subset elements in a set of n and the sum of them are even-thinking-(Fast Power modulo), hdu5363

HDU 5363 element is 1 ~ Number of subset elements in a set of n and the sum of them are even-thinking-(Fast Power modulo), hdu5363

Question: A set has element 1 ~ N, ask his subset to satisfy the condition that the sum of all elements in the subset is an even number and how many such subsets are there?

Analysis:

An arrangement and combination problem. If the element is an even number, the odd number must be an even number, and the even number does not matter. Therefore, the even number has a 2 ^ (n/2) method, multiply by the odd number (C (n + 1)/) + C (n + 1 ).....) method of selection, minus one, except for the empty set, note that when the odd number is obtained above, it is (n + 1)/2 (here is the downward Division ), is the combination of n as an even number and n as an odd number.

Number of combinations: 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 and even number)

Merge the formula (division here is the real division, so when n is an odd number, + 1 or-1 is used for Division ):

1. when n is an even number, there are 2 ^ (n/2) methods for obtaining an even number, and 2 ^ (n/N-1)/2 methods for obtaining an odd number) -1

2. when n is an odd number, the even number can be 2 ^ (n-1)/2), and the odd number can be 2 ^ (n + 1)/2, ans = 2 ^ (n-1)-1

The rest is the rapid power modulo. Remember this method.

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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