今日SGU 6.5

sgu 160

题意：给你n个数字 数字范围 1 到 m 问你从中取出任意数量的数字使得这些数字的积取模m最大

收获：dp，记录dp的路径

#include<bits/stdc++.h>
#define de(x) cout<<#x<<"="<<x<<endl;
#define dd(x) cout<<#x<<"="<<x<<" ";
#define rep(i,a,b) for(int i=a;i<(b);++i)
#define repd(i,a,b) for(int i=a;i>=(b);--i)
#define repp(i,a,b,t) for(int i=a;i<(b);i+=t)
#define ll long long
#define mt(a,b) memset(a,b,sizeof(a))
#define fi first
#define se second
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
#define pii pair<int,int>
#define pdd pair<double,double>
#define pdi pair<double,int>
#define mp(u,v) make_pair(u,v)
#define sz(a) (int)a.size()
#define ull unsigned long long
#define ll long long
#define pb push_back
#define PI acos(-1.0)
#define qc std::ios::sync_with_stdio(false)
#define db double
#define all(a) a.begin(),a.end()
const int mod = 1e9+7;
const int N = 1e4+6;
const double eps = 1e-6;
using namespace std;
bool eq(const db &a, const db &b) { return fabs(a - b) < eps; }
bool ls(const db &a, const db &b) { return a + eps < b; }
bool le(const db &a, const db &b) { return eq(a, b) || ls(a, b); }
ll gcd(ll a,ll b) { return a==0?b:gcd(b%a,a); };
ll lcm(ll a,ll b) { return a/gcd(a,b)*b; }
ll kpow(ll a,ll b) {ll res=1;a%=mod; if(b<0) return 1; for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll read(){
    ll x=0,f=1;char ch=getchar();
    while (ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//inv[1]=1;
//for(int i=2;i<=n;i++) inv[i]=(mod-mod/i)*inv[mod%i]%mod;
int a[N];
bool dp[2010];
pii pre[2010];
void out(int j){
//    de(j)
    if(!pre[j].fi) {
        printf("%d ",pre[j].se);
        return ;
    }
    out(pre[j].fi);
    printf("%d ",pre[j].se);
}
int main(){
    int n,m;
    scanf("%d%d",&n,&m);
    rep(i,1,n+1) scanf("%d",a + i),a[i] %= m;
    rep(i,1,n+1) rep(j,0,m) {
        if(dp[j]&&!dp[j*a[i]%m]&&pre[j].se!=i){
            dp[j*a[i]%m] = true; pre[j*a[i]%m] = mp(j,i);
        }
        if(!dp[a[i]]) dp[a[i]] = true,pre[a[i]] = mp(0,i);
    }
    repd(i,m,0) if(dp[i]) {
        if(i) printf("%d\n",i);
        out(i);
        return 0;
    }
    return 0;
}