hdu 4602 Partition

最新推荐文章于 2017-12-03 20:32:36 发布

原创最新推荐文章于 2017-12-03 20:32:36 发布 · 702 阅读

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本文介绍了一段使用C++解决组合数学问题的代码，包括幂运算取模、组合数计算等核心算法。

简单题

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<queue>
#include<stack>
#include<string>
#include<map>
#include<set>
#include<cmath>
#include<cassert>
#include<cstring>
#include<iomanip>
#include<ctime>
using namespace std;
#ifdef _WIN32
typedef __int64 i64;
#define out64 "%I64d\n"
#define in64 "%I64d"
#else
typedef long long i64;
#define out64 "%lld\n"
#define in64 "%lld"
#endif
/************ for topcoder by zz1215 *******************/
#define foreach(c,itr)  for(__typeof((c).begin()) itr=(c).begin();itr!=(c).end();itr++)
#define FOR(i,a,b)      for( int i = (a) ; i <= (b) ; i ++)
#define FF(i,a)         for( int i = 0 ; i < (a) ; i ++)
#define FFD(i,a,b)      for( int i = (a) ; i >= (b) ; i --)
#define S64(a)          scanf(in64,&a)
#define SS(a)           scanf("%d",&a)
#define LL(a)           ((a)<<1)
#define RR(a)           (((a)<<1)+1)
#define pb              push_back
#define pf              push_front
#define X               first
#define Y               second
#define CL(Q)           while(!Q.empty())Q.pop()
#define MM(name,what)   memset(name,what,sizeof(name))
#define MC(a,b)		memcpy(a,b,sizeof(b))
#define MAX(a,b)        ((a)>(b)?(a):(b))
#define MIN(a,b)        ((a)<(b)?(a):(b))
#define read            freopen("in.txt","r",stdin)
#define write           freopen("out.txt","w",stdout)

const int inf = 0x3f3f3f3f;
const long long inf64 = 0x3f3f3f3f3f3f3f3fLL;
const double oo = 10e9;
const double eps = 10e-9;
const double pi = acos(-1.0);
const int mod = 1000000007;

int n, k;

int pow2mod(int x)
{
	if (x < 0) return 0;
	int ans = 1;
	int temp = 2;
	while (x)
	{
		if (x & 1)
		{
			ans = ( (i64)ans*temp ) % mod; 
		}
		temp = ( (i64)temp*temp) % mod;
		x >>= 1;
	}
	return ans;
}

int gao(int n,int k)
{
	if (k > n) return 0;
	if (k == n) return 1;
	return  (pow2mod(n - k) + (((i64)(n - k - 1))*pow2mod(n - k - 2))%mod) % mod;
}

int main()
{
	int T;
	cin >> T;
	while (T--)
	{
		cin >> n >> k;
		cout << gao(n, k) << endl;
	}
	return 0;
}