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=100010;
const int INF=0x3f3f3f3f;
const int MOD= 1000000007,STA=8000010;
const LL LNF=1LL<<55;
const double EPS=1e-9;
const double OO=1e30;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//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=1;
    w[0]=0;
    for(i=1;w[k-1]<=100000;i++){
        w[k++]=(3*i*i-i)/2;
        w[k++]=(3*i*i+i)/2;
    }
    f[0]=1;
    for(i=1;i<=100000;i++){
        f[i]=0;
        for(j=1;w[j]<=i;j++){
            if(((j-1)>>1)&1)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
",(f[n]+MOD)%MOD);
    }
    return 0;
}