HDU- 6880 Permutation Counting (思维+dp)
题面:
题意:
给定一个整数\(\mathit n\),询问你有多少种\(1...n\) 的全排列\(\mathit a\)使其满足数组\(\mathit b\) 。
思路:
直接求排列不是很好求,我们做如下转化:
我们构造一个\([1,n]\) 的全排列\(\mathit p\),使其代表在\(\mathit a\) 排列中,第\(\mathit i\) 个数应该在位置\(\mathit p_i\)。
\(b_i=1\)时,代表右边的数大于左边的数,即:\(a_i>a_{i+1}\),我们只需要在排列\(\mathit p\) 中,\(i+1\)在\(\mathit i\) 的左边即可,这样就可以保证排列\(\mathit a\)中 \(i+1\) 位置上的数字一定小于\(\mathit i\) 位置上的数字。
而满足条件的排列\(\mathit p\) 的数量我们时可以\(dp\) 求出的,
我们设\(dp[i][j]\)代表处理到数字\(\mathit i\)时,且\(\mathit i\) 在位置\(\mathit j\) 上的方案数。
转移:
\(b_i=1,dp[i][j]=\sum_{k=j}^{k=i}dp[i-1][k]\\b_i=0,dp[i][j]=\sum_{k=1}^{k=j-1}dp[i-1][k]\)
代码:
using namespace std;
typedef long long ll;
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
ll lcm(ll a, ll b) {return a / gcd(a, b) * b;}
ll powmod(ll a, ll b, ll MOD) { if (a == 0ll) {return 0ll;} a %= MOD; ll ans = 1; while (b) {if (b & 1) {ans = ans * a % MOD;} a = a * a % MOD; b >>= 1;} return ans;}
ll poww(ll a, ll b) { if (a == 0ll) {return 0ll;} ll ans = 1; while (b) {if (b & 1) {ans = ans * a ;} a = a * a ; b >>= 1;} return ans;}
void Pv(const vector<int> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%d", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
void Pvl(const vector<ll> &V) {int Len = sz(V); for (int i = 0; i < Len; ++i) {printf("%lld", V[i] ); if (i != Len - 1) {printf(" ");} else {printf("\n");}}}
inline long long readll() {long long tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
inline int readint() {int tmp = 0, fh = 1; char c = getchar(); while (c < '0' || c > '9') {if (c == '-') fh = -1; c = getchar();} while (c >= '0' && c <= '9') tmp = tmp * 10 + c - 48, c = getchar(); return tmp * fh;}
void pvarr_int(int *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%d%c", arr[i], i == n ? '\n' : ' ');}}
void pvarr_LL(ll *arr, int n, int strat = 1) {if (strat == 0) {n--;} repd(i, strat, n) {printf("%lld%c", arr[i], i == n ? '\n' : ' ');}}
const int maxn = 5000 + 10;
const int inf = 0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
ll dp[maxn][maxn];
int a[maxn];
const ll mod = 1e9 + 7ll;
int main()
{
freopen("C:\\code\\input.txt", "r", stdin);
//freopen("C:\\code\\output.txt","w",stdout);
int t;
t = readint();
while (t--)
{
int n = readint();
repd(i, 2, n)
{
a[i] = readint();
}
repd(i, 1, n)
{
repd(j, 1, n)
{
dp[i][j] = 0;
}
}
dp[1][1] = 1;
ll sum;
repd(i, 2, n)
{
if (a[i] == 1)
{
sum = 0ll;
repd(j, 1, i)
{
sum = (sum + dp[i - 1][j]) % mod;
}
repd(j, 1, i)
{
dp[i][j] = (dp[i][j] + sum) % mod;
sum = (sum - dp[i - 1][j] + mod) % mod;
}
} else
{
sum = 0ll;
repd(j, 1, i)
{
dp[i][j] = (dp[i][j] + sum) % mod;
sum = (sum + dp[i - 1][j]) % mod;
}
}
}
ll ans = 0ll;
repd(i, 1, n)
{
ans = (ans + dp[n][i]) % mod;
}
printf("%lld\n", ans );
}
return 0;
}