牛客练习赛52 烹饪(裴蜀定理+dp)

  题目大意就是给你n个数,然后要找这n个数的线性组合a1*x1+a2*x2+a3*x3+....+am*xm 等于任意正整数(1<=m<=n)，问你这样的组合（a1,a2,...,am）有多少个，因为a1*x1+a2*x2+a3*x3+....+am*xm 等于任意正整数 ，然后任意正整数都是1的倍数，所以由裴蜀定理可知，选取的那m个数的gcd要为1才行，然后题目就转化为取若干个数gcd为1的方案数了，可以dp求，用的是刷表法

#include<bits/stdc++.h>
using namespace std;
#define ls rt<<1
#define rs (rt<<1)+1
#define PI acos(-1)
#define eps 1e-8
#define ll long long
#define fuck(x) cout<<#x<<"     "<<x<<endl;
typedef pair<int,int> pii;
const int inf=2e9;
const int maxn=1e6+10;
int d[4][2]={1,0,-1,0,0,1,0,-1};
//int lowbit(int x){return x&-x;}
//void add(int x,int v){while(x<=n)bit[x]+=v,x+=lowbit(x);}
//int sum(int x){int ans=0;while(x>=1) ans+=bit[x],x-=lowbit(x);return ans;}
inline ll read() {
    ll s = 0,w = 1;
    char ch = getchar();
    while(!isdigit(ch)) {
        if(ch == '-') w = -1;
        ch = getchar();
    }
    while(isdigit(ch))
        s = s * 10 + ch - '0',ch = getchar();
    return s * w;
}
inline void write(ll x) {
    if(x < 0)
        putchar('-'), x = -x;
    if(x > 9)
        write(x / 10);
    putchar(x % 10 + '0');
}
int mod=998244353;
int dp[3005][2005],a[3005];
int gcd(int x,int y){
    return y==0?x:gcd(y,x%y);
}
int main(){
    int n;
    n=read();
    for(int i=1;i<=n;i++)
        a[i]=read();
    dp[1][a[1]]=1;
    dp[1][0]=1;
    for(int i=1;i<n;i++)
        for(int j=0;j<=2000;j++)
        {
            dp[i+1][j]=(dp[i+1][j]+dp[i][j])%mod;
            dp[i+1][gcd(j,a[i+1])]=(dp[i+1][gcd(j,a[i+1])]+dp[i][j])%mod;
        }
    write(dp[n][1]);puts("");
    return 0;
}