uva 10581 - Partitioning for fun and profit(记忆化搜索+数论)

题目链接：uva 10581 - Partitioning
for fun and profit

题目大意：给定m，n，k，将m分解成n份，然后按照每份的个数排定字典序，并且划分时要求ai?1≤ai，然后输出字典序排在k位的划分方法。

解题思路：因为有ai?1≤ai的条件，所以先记忆化搜索处理出组合情况dp[i][j][s]表示第i位为j，并且剩余的未划分数为s的总数为dp[i][j][s],然后就是枚举每一位上的值，判断序列的位置即可。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 220;
const int maxp = 10;

int M, N;
ll K, dp[maxp+5][maxn+5][maxn+5];

ll dfs (int d, int x, int s) {
    if (d == N) {
        if (0 == s)
            return 1;
        else
            return 0;
    }

    ll& ans = dp[d][x][s];

    if (ans != -1)
        return ans;

    ans = 0;

    for (int i = x; ; i++) {
        if ((N-d) * i > s)
            break;
        ans += dfs(d+1, i, s-i);
    }
    return ans;
}

void solve () {
    int s = M, t = 1;
    for (int i = 1; i < N; i++) {
        for (int j = t; ; j++) {
            ll u = dp[i][j][s-j];

            if (K > u) {
                K -= u;
            } else {
                printf("%d\n", j);
                s -= j;
                t = j;
                break;
            }
        }
    }
    printf("%d\n", s);
}

int main () {
    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d%d%lld", &M, &N, &K);

        memset(dp, -1, sizeof(dp));
        for (int i = 1; i * N <= M; i++)
            ll u = dfs(1, i, M-i);
        solve();
    }
    return 0;
}

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