计蒜客 划分整数

$dp(i,j,lim),(j \le i)$表示将$i$划分为不超过$j$个正整数且最大正整数不超过$lim$的方案数。

AC代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 300 + 5;
LL dp[maxn][maxn][maxn];

LL solve(int n, int k, int lim) {
    if(k*lim < n) return 0;
    if(n == 0 && k == 0) return 1;
    if(dp[n][k][lim] != -1) return dp[n][k][lim];
    dp[n][k][lim] = 0;
    for(int i = 1; i <= lim; i++) {
        dp[n][k][lim] += solve(n-i, min(n-i, k-1), i);
    }
    return dp[n][k][lim];
}

int main() {
    int n, k;
    while(scanf("%d%d", &n, &k) == 2) {
        memset(dp, -1, sizeof(dp));
        printf("%lld\n", solve(n, min(n,k), n));
    }
    return 0;
}

如有不当之处欢迎指出！