NYOJ 571 —— 各种划分数

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
using namespace std;

int dp[55][55];

int main ()
{
    int n, k, a, b, c, d, e;
    while(scanf("%d%d", &n, &k) != EOF) {
        // 把正整数i划分为不超过j的正整数的划分个数
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=n; j++) {
                if(j>i)    dp[i][j] = dp[i][i];
                else if(j == i)    dp[i][j] = dp[i][j-1] + 1;
                else    dp[i][j] = dp[i][j-1] + dp[i-j][j];
            }
        }
        a = dp[n][n];
        c = dp[n][k];
        
        // 把正整数划分为k个正整数的划分个数
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=k; j++) {
                if(j>i)    dp[i][j] = 0;
                else if(j==i)    dp[i][j] = 1;
                else    dp[i][j] = dp[i-1][j-1] + dp[i-j][j];
            }
        }
        b = dp[n][k];
        
        
        // 把i划分为不超过j的奇数的划分个数
        int odd = n&1 ? n : n-1;
        for(int i=1; i<=n; i++) {
            dp[i][1] = 1;
            for(int j=3; j<=odd; j+=2) {
                if(j>i) {
                    if(i&1)    dp[i][j] = dp[i][i];
                    else    dp[i][j] = dp[i][i-1];
                }
                else if(j==i)    dp[i][j] = dp[i][i-2] + 1;
                else    dp[i][j] = dp[i][j-2] + dp[i-j][j];
            }
        }
        d = dp[n][odd];
        
        // 把i划分为不同正整数的划分个数
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=n; j++) {
                if(j>i)    dp[i][j] = dp[i][i];
                else if(j == i)    dp[i][j] = dp[i][i-1] + 1;
                else    dp[i][j] = dp[i][j-1] + dp[i-j][j-1];
            }
        }
        e = dp[n][n];
              
        printf("%d\n%d\n%d\n%d\n%d\n\n", a, b, c, d, e);
        memset(dp, 0, sizeof(dp));
    }
    
    return 0;
}