swust_wbh的专栏

题目在http://acm.hdu.edu.cn/showproblem.php?pid=4542  当type=0的时候，就是求因子个数为k的最小数。我们知道，一个数可以这样分解：  其中p1,p2……,pr都是素数并且都不相同所以思路就是将k这个数进行分解，然后搜索出所有的情况，找到最小值即可。当type=1的时候，设我们求的数为n，那么这个数的因子个数就为n-k;求这样一

先考虑N!最后一位非0的数的求法。我们构造一个数组table[4]={6,2,4,8}，这个数组的特点就是table[(i+1)%4]=(table[i]*2)%10那么table[i]*2相当于是数组下标往右移动了一位（移动到最右边就从左边循环），table[i]*1  不变table[i]*2  右1位table[i]*3  右3位table[i]*4

若x=a0^p0*a1^p1*……*ai^pi(a0,a1……,ai是不同的素数)对于条件3：因子和=(a0^0+a0^1+……+a0^p0)*(a1^0+a1^1+……+a1^p1)*……如果是素数，显然质因子个位要为1，并且(a0^(p0+1)-1)/(a0-1)为素数对于条件4：设因子积为a0^q0*a1^q1*……*ai^qi(a0,a1……,ai是不同的素数)则有qi=(

a[1]=1;a[2]=2;a[n]=2*a[n-1]+2^(n-3)=2^2*a[n-2]+2*2^(n-3)=2^3[n-3]+3*2^(n-3)=……=2^(n-2)a[2]+(n-2)*2^(n-3)=2^(n-1)+(n-2)*2^(n-3);#include#include#include#include#include#includeusing namespa

#include#include#include#include#include#include#includeusing namespace std;#define lc l,m,index#define rc m+1,r,index#define ll long long#define inf 0x3f3f3f3f#define N 100000

#include#include#include#include#include#include#includeusing namespace std;#define lc l,m,index<<1#define rc m+1,r,index<<1|1#define ll long long#define inf 0x3f3f3f3f#define N 100005#de

#include#include#include#include#include#include#includeusing namespace std;#define lc l,m,index<<1#define rc m+1,r,index<<1|1#define ll long long#define inf 0x3f3f3f3f#define N 1000005#d

#include#include#include#include#include#includeusing namespace std;#define lc l,m,index<<1#define rc m+1,r,index<<1|1#define ll long long#define inf 0x3f3f3f3f#define N 10005int n,m,ans;

#include#include#include#include#include#includeusing namespace std;#define lc l,m,index<<1#define rc m+1,r,index<<1|1#define ll long long#define inf 0x3f3f3f3f#define N 1000005ll n;ll ph

#include#include#include#include#include#includeusing namespace std;#define lc l,m,index<<1#define rc m+1,r,index<<1|1#define ll long long#define inf 0x3f3f3f3fll n,a;bool prime(ll num){

#include#include#include#include#include#include#includeusing namespace std;#define lc l,m,index<<1#define rc m+1,r,index<<1|1#define ll long long#define inf 0x3f3f3f3f#define N 100005#de

构造An=(a+sqrt(b)^n+(a-sqrt(b))^n，An为整数。由于b的范围，求An即可。An*((a+sqrt(b))+(a-sqrt(b)))=...=2*a*An=...=A(n+1)+(a*a-b)*A(n-1),得到递推关系A(n+1)=2*a*An+(b-a*a)*A(n-1)。构造矩阵temp[2][2]={2*a,-a*a+b,1,0}.使用矩阵快速

设f(0)=1,f(1)=0,f(n)=f(n-1)+f(n-2)(n>=2)。那么有F(n)=a^f(n)*b^f(n+1)。根据费马小定理，1000000007是一个素数，那么所求可以化简为a^(f(n)%1000000006)*b^(f(n+1)%1000000006),那么就可以先求f(n)和f(n+1)，再用二分快速幂求得结果。由于n比较大，求f(n)和f(n+1)要用矩

#include#include#include#include#includeusing namespace std;#define ll long long#define N 2struct matrix{ll m[N][N];};ll ans[N][N],temp[N][N];matrix mul(ll a[N][N],ll b[N][N]

#include#include#include#include#include#include#includeusing namespace std;#define N 25#define ll __int64ll dp[N][3];ll n;//dp[i][0] 含49的//dp[i][1] 不含49且最高位是9的//dp[i][2] 不含49且最高位不是9的