hdu 1028 Ignatius and the Princess III 简单dp

题目链接：hdu 1028 Ignatius and the Princess III 

题意：对于给定的n，问有多少种组成方式

思路：dp[i][j]，i表示要求的数，j表示组成i的最大值，最后答案是dp[i][i]。那么dp[i][j]=dp[i][j-1]+dp[i-j][i-j]，dp[i][j-1]是累加1到j-1的结果，dp[i-j][i-j]表示的就是最大为j，然后i-j有多少种表达方式啦。因为i-j可能大于j，这与我们定义的j为最大值矛盾，所以要去掉大于j的那些值

/**************************************************************
    Problem:hdu 1028
    User: youmi
    Language: C++
    Result: Accepted
    Time:15MS
    Memory:1908K
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%I64d",&a)
#define pt(a) printf("%d
",a)
#define ptlld(a) printf("%I64d
",a)
#define rep(i,from,to) for(int i=from;i<=to;i++)
#define irep(i,to,from) for(int i=to;i>=from;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define eps 1e-6
#define oo 0x3fffffff
#define TEST cout<<"*************************"<<endl
const double pi=4*atan(1.0);

using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
    char c; int flag = 1;
    for (c = getchar(); !(c >= '0' && c <= '9' || c == '-'); c = getchar()); if (c == '-') flag = -1, n = 0; else n = c - '0';
    for (c = getchar(); c >= '0' && c <= '9'; c = getchar()) n = n * 10 + c - '0'; n *= flag;
}
int Pow(int base, ll n, int mo)
{
    if (n == 0) return 1;
    if (n == 1) return base % mo;
    int tmp = Pow(base, n >> 1, mo);
    tmp = (ll)tmp * tmp % mo;
    if (n & 1) tmp = (ll)tmp * base % mo;
    return tmp;
}
//***************************

int n;
const int maxn=200+10;
ll dp[maxn][maxn];
void init()
{
    zeros(dp);
    dp[0][0]=1;
    rep(i,1,120)
    {
        rep(j,1,i)
        {
            dp[i][j]=dp[i][j-1]+dp[i-j][i-j];
            if(j<(i+1)/2)
                dp[i][j]-=dp[i-j][i-j]-dp[i-j][j];
        }
    }
}

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif
    init();
    while(~sc(n))
    {
        ptlld(dp[n][n]);
    }
}