sgu269：Rooks(dp)

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题目大意：
给出n段长的为整数ai的格子，按照左端点对齐。在其中放车，每两个车不能同时在同一行或同一列。求出放入k个车的方案总数。

分析：
把ai按照从小到大排序，长的小的段放一个一定会影响后面的所有段，因此考虑fi,j表示前i行放了j个车的方案数，方程为fi,j=fi−1,j+(a[i]−j+1)fi−1,j−1。
要加高精度。

AC code：

#include <cstdio>#include <cmath>#include <cstdlib>#include <cstring>#include <cctype>#include <algorithm>#include <string>#include <sstream>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <queue>#include <vector>#define pb push_back#define mp make_pairtypedef long long LL;typedef double DB;typedef long double LD;using namespace std;const int MAXN = 259;const int MAXL = 209;const int mod = 10000;int n, k;int a[MAXN];struct BigNum{    int len;    int a[MAXL];    BigNum()     {        len = 1;        memset(a, 0, sizeof a);     }    void init(int x) {a[len = 1] = x;}    void shift()    {        for(int i = 1; i <= len; ++i)        {            a[i+1] += a[i]/mod;            a[i] %= mod;            }        while(a[len+1])        {            len++;            a[len+1] += a[len]/mod;            a[len] %= mod;        }        while(!a[len] && len) len--;        if(!len) len = 1;    }    friend BigNum operator + (const BigNum &a, const BigNum &b)    {        BigNum ret;        int maxl = max(a.len, b.len);        for(int i = 1; i <= maxl; ++i)            ret.a[i] += a.a[i]+b.a[i];        ret.len = maxl;        ret.shift();        return ret;    }    friend BigNum operator * (const BigNum &a, int k)    {        BigNum ret = a;        for(int i = 1; i <= ret.len; ++i)            ret.a[i] *= k;        ret.shift();        return ret;     }       void print()    {        printf("%d", a[len]);        for(int i = len-1; i >= 1; --i)            printf("%04d", a[i]);       }};BigNum f[MAXN];int main(){    #ifndef ONLINE_JUDGE    freopen("input.txt", "r", stdin);    freopen("output.txt", "w", stdout);    #endif    scanf("%d%d", &n, &k);    for(int i = 1; i <= n; ++i)        scanf("%d", a+i);    sort(a+1, a+n+1);    for(int i = 0; i <= n; ++i) f[0].init(1);    for(int i = 1; i <= n; ++i)        for(int j = a[i]+1; j >= 1; --j)            f[j] = f[j]+f[j-1]*(a[i]-j+1);    f[k].print();    #ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);    #endif    return 0;}

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