BZOJ2655 Calc

BZOJ2655 Calc

参考

题意：

　　给定n，m，mod，问在对mod取模的背景下，从【1，m】中选出n个数相乘可以得到的总和为多少。

思路：

　　首先可以发现dp方程 ，假定dp【m】【n】表示从【1 ~ m】中选出n个数乘积的和，

那么dp【m】【n】 = dp【m-1】【n】 + dp【m-1】【n-1】*m*n。

但是这道题的m有1e9那么大，不能dp完，不过我们可以发现，dp【x】【n】 是关于x的2*n多项式，

所以，我们只要先求出0~2*n的dp值，再用拉格朗日插值法算出dp【m】【n】的即可。

#include <algorithm>
#include  <iterator>
#include  <iostream>
#include   <cstring>
#include   <iomanip>
#include   <cstdlib>
#include    <cstdio>
#include    <string>
#include    <vector>
#include    <bitset>
#include    <cctype>
#include     <queue>
#include     <cmath>
#include      <list>
#include       <map>
#include       <set>
using namespace std;
//#pragma GCC optimize(3)
//#pragma comment(linker, "/STACK:102400000,102400000")  //c++
#define lson (l , mid , rt << 1)
#define rson (mid + 1 , r , rt << 1 | 1)
#define debug(x) cerr << #x << " = " << x << "
";
#define pb push_back
#define pq priority_queue



typedef long long ll;
typedef unsigned long long ull;

typedef pair<ll ,ll > pll;
typedef pair<int ,int > pii;
typedef pair<int ,pii> p3;
//priority_queue<int> q;//这是一个大根堆q
//priority_queue<int,vector<int>,greater<int> >q;//这是一个小根堆q
#define fi first
#define se second
//#define endl '
'

#define OKC ios::sync_with_stdio(false);cin.tie(0)
#define FT(A,B,C) for(int A=B;A <= C;++A)  //用来压行
#define REP(i , j , k)  for(int i = j ; i <  k ; ++i)
//priority_queue<int ,vector<int>, greater<int> >que;

const ll mos = 0x7FFFFFFFLL;  //2147483647
const ll nmos = 0x80000000LL;  //-2147483648
const int inf = 0x3f3f3f3f;
const ll inff = 0x3f3f3f3f3f3f3f3fLL; //18

const double PI=acos(-1.0);

template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0'||ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
// #define _DEBUG;         //*//
#ifdef _DEBUG
freopen("input", "r", stdin);
// freopen("output.txt", "w", stdout);
#endif
/*-----------------------show time----------------------*/

            const int maxn = 3000;
            ll dp[maxn][maxn],x[maxn],y[maxn];
            int m,n,mod;

            ll ksm (ll a,ll b){
                ll res = 1;
                while(b>0){
                    if(b&1) res = (res * a)%mod;
                    a = (a * a)%mod;
                    b >>= 1;
                }
                return res;
            }
            ll lagerange(int k){
                ll res = 0;
                for(int i=0; i<=2*n; i++){
                    ll s1=1,s2 = 1;

                    for(int j=0; j<=2*n; j++){
                            if(i==j)continue;
                            s1 = 1ll*(s1 * (k - x[j] + mod)%mod)%mod;
                            s2 = 1ll*(s2 * ((x[i] - x[j] + mod)%mod))%mod;
                    }
                    res = (res + 1ll*s1 * ksm(s2,mod-2) % mod * y[i] % mod+mod)%mod;
                }
                return res;
            }
int main(){
            
            scanf("%d%d%d", &m, &n, &mod);
            dp[0][0] = 1;
            for(int i=1; i<=2*n; i++){
                dp[i][0] = 1;
                for(int j=1; j<=n; j++){
                    dp[i][j] = 1ll*dp[i-1][j-1] * i % mod * j + dp[i-1][j];
                    dp[i][j] = dp[i][j]%mod;
                }
            }

            if(m <= 2 * n){
                printf("%lld
", dp[m][n]);
                return 0;
            }

            for(int i=1; i<=2*n; i++) x[i] = i,y[i] = dp[i][n];

            printf("%lld
",lagerange(m));


            return 0;
}