CF-431-D-二分+数位DP

合集 - Codeforces(31)

1.CF-55-D-数位DP2024-01-232.CF-87-D-并查集2024-01-133.CF-91-B-单调栈+二分2024-01-234.CF-240-F-线段树2024-01-215.CF-242-E-线段树2024-01-206.CF-292-D-并查集2024-01-237.CF-282-E-Trie2024-01-178.CF-342-E-分块2024-01-129.CF-375-D-启发式合并2024-01-12

10.CF-431-D-二分+数位DP2024-01-24

11.CF-461-B-树形DP2024-01-2212.CF-514-D-单调队列2024-01-1413.CF-522-D-线段树2024-05-0514.CF-570-D-启发式合并2024-01-2115.CF-600-E-启发式合并2024-05-0416.CF-613-D-虚树2024-01-1317.CF-620-E-DFS序+线段树2024-01-1718.CF-675-E-RMQ优化DP2024-01-1619.CF-702-E-倍增2024-01-1720.CF-797-E-根号分治2024-05-0421.CF-817-E-Trie2024-01-1622.CF-877-E-dfs序+线段树2024-05-0523.CF-915-F-并查集2024-01-2124.CF-938-D-最短路2024-05-1025.CF-1051-F-最短路+最小生成树2024-01-1726.CF-1184-E3-最小生成树+倍增+并查集2024-01-2427.CF-1304-E-倍增LCA+思维2024-01-2528.CF-1399-E2-优先队列2024-01-2229.CF-1515-F-思维2024-02-2430.CF-1831-E-卡特兰数+异或哈希+差分2024-01-2231.CF-1921-F-根号分治2024-01-24

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431-D 题目大意

请你找到一个数$n$，满足区间$[n+1,2n]$中恰有$m$个数的二进制表示中有$k$个$1$。

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Solution

这种区间中计数类型的题目首先相当数位DP。

但是这里缺乏上下界，难点就在于观察到$n$的单调性($[n+1,2n]$中有$k$个$1$的数是单调不减的)，简要证明：

• 对于一个$n$对应的数为$n+1,n+2,····,2n-1,2n$。
• 而对于$n+1$对应的数为$n+2,n+3,····,2n-1,2n,2n+1,2n+2$。

这里面前者多了个$n+1$，后者多了$2n+1,2n+2$两个数，其余数都一样，这里$n+1$和$2n+2$的二进制表示中$1$的个数相同，因为他们是两倍的关系，那么相当于后者多出来一个$2n+1$，因此具有单调性。

二分答案$n$，数位DP计数，时间复杂度$O(lo{g}^{2}U)$。

#include<bits/stdc++.h>
using namespace std;
using ll=long long;

ll f[66][66];

ll calc(ll mid,int k){
    ll l=mid,r=mid*2;
    vector<int> num(64);
    function<ll(int,int,int)> dp=[&](int pos,int full,int cnt)->ll{
        if(pos<0) return cnt==k;
        if(pos>=num.size()) return dp(pos-1,full,cnt);
        if(!full&&f[pos][cnt]!=-1) return f[pos][cnt];
        ll res=0;
        for(int i=0;i<=1;i++){
            if(full&&i>num[pos]) break;
            int nxtfull=(full&&i==num[pos]);
            int nxtcnt=cnt+i;
            if(nxtcnt<=k) res+=dp(pos-1,nxtfull,nxtcnt);
        }
        if(!full) f[pos][cnt]=res;
        return res;
    };
    for(int i=63;~i;i--){
        if(l&(1LL<<i)) num[i]=1;
        else num[i]=0;
    }
    ll low=dp(63,1,0);
    for(int i=63;~i;i--){
        if(r&(1LL<<i)) num[i]=1;
        else num[i]=0;
    }
    ll high=dp(63,1,0);
    return high-low;
}

void solve(){
    ll m,k;
    cin>>m>>k;
    ll l=1,r=1e18;
    memset(f,-1,sizeof f);
    while(l<r){
        ll mid=(l+r)>>1;
        if(calc(mid,k)<m) l=mid+1;
        else r=mid;
    }
    cout<<l<<'\n';
}

int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    //freopen("input.txt","r",stdin);
    //freopen("output.txt","w",stdout);
    int T=1;
    //cin>>T;
    while(T--){
        solve();
    }
    return 0;
}