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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