hdu 4602 Partition (递推+二分快速幂）

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找规律，可得到

n 1 2 3 4 5 .......

k =1 1 2 5 12 28 .......

=2 0 1 2 5 12 .......

=3 0 0 1 2 5 .......

=4 0 0 0 1 2 .......

=5 0 0 0 0 1 .......

推得：f[n]=f[n-1]*2+2^(n-3) ------> f[n]=(n-2)*2^(n-3)+2^(n-1)

#include<iostream>#include<cstring>#include<string>#include<cstdio>using namespace std;const long long modd=1000000007;long long T,n,k,ans;long long power(long long x,long long y){    long long ans=1,k=x;    while(y>0)    {       if(y & 1) ans=(ans*k)% modd;       y>>=1;       k=(k*k)% modd;    }    return ans;}int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&n,&k);        n=n-(k-1);        if(n<=0) cout<<0<<endl;        else if(n==1) cout<<1<<endl;        else if(n==2) cout<<2<<endl;        else        {            k=power(2,n-3) % modd;            ans=((n-2)*k+k*4) % modd;            cout<<ans<<endl;        }    }    return 0;}