HDU-4651 Partition 整数拆分,递推

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4651

　　题意：求n的整数拆为Σ i 的个数。

　　一般的递归做法，或者生成函数做法肯定会超时的。。。

　　然后要奇葩的用到一个<五边形数定理>,然后根据公式递推就可以了，先预处理下，复杂度O(n*sqrt(n))..

 //STATUS:C++_AC_796MS_1012KB

 #include <functional>

 #include <algorithm>

 #include <iostream>

 //#include <ext/rope>

 #include <fstream>

 #include <sstream>

 #include <iomanip>

 #include <numeric>

 #include <cstring>

 #include <cassert>

 #include <cstdio>

 #include <string>

 #include <vector>

 #include <bitset>

 #include <queue>

 #include <stack>

 #include <cmath>

 #include <ctime>

 #include <list>

 #include <set>

 #include <map>

 using namespace std;

 //#pragma comment(linker,"/STACK:102400000,102400000")

 //using namespace __gnu_cxx;

 //define

 #define pii pair<int,int>

 #define mem(a,b) memset(a,b,sizeof(a))

 #define lson l,mid,rt<<1

 #define rson mid+1,r,rt<<1|1

 #define PI acos(-1.0)

 //typedef

 typedef __int64 LL;

 typedef unsigned __int64 ULL;

 //const

 const int N=;

 const int INF=0x3f3f3f3f;

 const int MOD= ,STA=;

 const LL LNF=1LL<<;

 const double EPS=1e-;

 const double OO=1e30;

 const int dx[]={-,,,};

 const int dy[]={,,,-};

 const int day[]={,,,,,,,,,,,,};

 //Daily Use ...

 inline int sign(double x){return (x>EPS)-(x<-EPS);}

 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}

 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}

 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}

 template<class T> inline T Min(T a,T b){return a<b?a:b;}

 template<class T> inline T Max(T a,T b){return a>b?a:b;}

 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}

 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}

 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}

 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}

 //End

 int w[N];

 LL f[N];

 int T,n;

 int main(){

  //   freopen("in.txt","r",stdin);

     int i,j,k=;

     w[]=;

     for(i=;w[k-]<=;i++){

         w[k++]=(*i*i-i)/;

         w[k++]=(*i*i+i)/;

     }

     f[]=;

     for(i=;i<=;i++){

         f[i]=;

         for(j=;w[j]<=i;j++){

             if(((j-)>>)&)f[i]=(f[i]-f[i-w[j]])%MOD;

             else f[i]=(f[i]+f[i-w[j]])%MOD;

         }

     }

     scanf("%d",&T);

     while(T--)

     {

         scanf("%d",&n);

         printf("%I64d\n",(f[n]+MOD)%MOD);

     }

     return ;

 }