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 ;
 }