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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;
}
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hdu 5225 Tom and permutation(回溯)
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原文地址:http://blog.csdn.net/keshuai19940722/article/details/47115903