[QOJ 6355] 5

發現至少有 \(\frac{S}{5}\) 個 \(1\)，所以考慮維護 \((k,T-k)\) 的對數，然後二進位制拆分+維護區間連續段揹包 dp 即可。

點選檢視程式碼

#include<bits/stdc++.h>
#define fir first
#define sec second
#define int long long
#define lowbit(x) x&(-x)
#define mkp(a,b) make_pair(a,b)
using namespace std;
typedef pair<int,int> pir;
inline int read(){
	int x=0,f=1; char c=getchar();
	while(!isdigit(c)){if(c=='-') f=-1; c=getchar();}
	while(isdigit(c)){x=x*10+(c^48); c=getchar();}
	return x*f;
}
const int inf=1e18,N=2e5+5;
int n,S;
int a[N],cnt[N],op,ans;
vector<pir> dp[N*2];
inline int P(int x){return x+n;}
inline void merge(vector<pir> &x,int L,int R){
    if(x.empty()){x.push_back(mkp(L,R));return ;}
    vector<pir> y; y.clear();
    for(auto [l,r]:x){
        if(r<L||R<l){y.push_back(mkp(l,r));continue;}
        L=min(L,l),R=max(R,r);
    }
    y.push_back(mkp(L,R));
    swap(x,y);
}
inline void upd(int val,int siz){
    if(val>0){
        for(int i=S;i>=-op;i--){
            if(dp[P(i)].empty()) continue;
            for(auto [l,r]:dp[P(i)])
            merge(dp[P(i+val)],l+siz,r+siz);
        }   
    }
    else{
         for(int i=-op;i<=S;i++){
            if(dp[P(i)].empty()) continue;
            for(auto [l,r]:dp[P(i)])
            merge(dp[P(i+val)],l+siz,r+siz);
        }
    }
}
signed main(){
    n=read(),S=read();
    for(int i=1;i<=n;i++) a[i]=read(),cnt[a[i]]++;
    dp[P(0)].push_back(mkp(0,cnt[1]));
    op=cnt[0];
    for(int i=0;i<=S;i++){
        if(!cnt[i]||i==1) continue;
        for(int w=1;w<=cnt[i];w<<=1){
            int val=i*w-w;
            cnt[i]-=w;
            upd(val,w);
        }
        if(cnt[i]) upd(cnt[i]*i-cnt[i],cnt[i]);
    }
    for(int i=-op;i<=S;i++){
        for(auto [l,r]:dp[P(i)]) ans+=(r-l+1);
    }
    cout<<ans<<'\n';
}