hdu 5225 Tom and permutation（回溯）

题目链接：hdu 5225 Tom and permutation

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

using namespace std;
typedef long long ll;

const int maxn = 100;
const int mod = 1e9+7;
int N, ans, V[maxn + 5], A[maxn + 5];
ll S[maxn + 5], L[maxn + 5];

int query(int x) {
int ret = 0;
for (int i = 1; i < x; i++) {
if (V[i] == 0)
ret++;
}
return ret;
}

int dfs(int d, int s) {

if (d > N)
return 0;

if (s == 0) {
ans = (ans + S[N-d+1]) % mod;
return L[N-d+1];
} else {

int ret = 0;

for (int i = 1; i <= A[d]; i++) {
if (V[i]) continue;

V[i] = 1;

int s = query(i);
ll temp = dfs(d + 1, i == A[d] ? 1 : 0);
ans = (ans + temp * s % mod) % mod;

ret = (ret + temp) % mod;

V[i] = 0;
}

return ret;
}
}

int main () {
S[1] = 0;
L[1] = 1;
for (int i = 2; i <= maxn; i++) {
S[i] = S[i-1] * i % mod+ L[i-1] * ((1LL * i * (i-1) / 2) % mod) % mod;
L[i] = L[i-1] * i % mod;
}

while (scanf("%d", &N) == 1) {
ans = 0;
memset(V, 0, sizeof(V));

for (int i = 1; i <= N; i++)
scanf("%d", &A[i]);

dfs(1, 1);
printf("%d\n", ans);
}
return 0;
}