hdu 5482 Numquam vincar(暴力)

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题目链接：hdu 5482 Numquam vincar

解题思路

dp[i][j][k]表示到长度为i，使用颜色为j，不相同子串有k个时的方案数，求dp数组的过程用回溯暴力一下。剩下就是对应乘上组合数。

代码

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 15;const int maxm = 105;const int mod = 1e9 + 7;const int base = 23;typedef long long ll;int arr[maxn], dp[maxn][maxn][maxm];ll temp[maxn];void modify(int d, int c) {    int n = 0;    for (int i = 1; i <= d; i++) {        ll s = 0;        for (int j = i; j <= d; j++) {            s = s * base + arr[j];            temp[n++] = s;        }    }    sort(temp, temp + n);    int k = unique(temp, temp + n) - temp;    dp[d][c][k]++;}void dfs (int d, int c) {    if (d > 10) return ;    for (int i = 1; i <= c; i++) {        arr[d] = i;        modify(d, max(i, c-1));        dfs(d+1, max(i+1, c));    }}int N, M, K;int solve () {    ll c = 1;    int ret = 0;    for (int i = 1; i <= N; i++) {        c = c * (K-i+1) % mod;        ret = (ret + c * dp[N][i][M] % mod) % mod;    }    return ret;}int main () {    memset(dp, 0, sizeof(dp));    dfs(1, 1);    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%d%d%d", &N, &M, &K);        printf("%d\n", solve());    }    return 0;}

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