題目鏈接

BZOJ4245

題解

套路①
位運算當然要分位討論，高位優先
考慮在\(or\)下，如果該位為\(0\)，則每一位都為\(0\)

套路②
我們選m段異或和，轉化為\(m\)個前綴和的點，且其中有一個是\(n\)

容易發現，該位結果要為0，則選取的前綴和該位都為\(0\)
所以貪心查找所有該位為\(0\)的，首先第\(n\)個前綴和一定要為\(0\)，如果其它滿足有至少\(m - 1\)個，那麽該位答案為\(0\)，剩余為\(1\)的打上標記不能選
如果不夠，那這一位就沒辦法了，直接放棄

#include<algorithm>
#include<iostream>
#include<cstring>
 

#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
 

using namespace std;
const int maxn = 500005,maxm = 100005,INF = 1000000000;
inline LL read(){
    LL out = 0,flag = 1; char c = getchar();
    while (c < 48 || c > 57){if (c == ‘-‘) flag = -1; c = getchar();}
    while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
    return
 
 out * flag;
}
int n,m,tag[maxn];
LL a[maxn],ans;
int main(){
    n = read(); m = read();
    REP(i,n) a[i] = read() ^ a[i - 1];
    for (LL i = 1ll << 61; i; i >>= 1){
        if (a[n] & i){
            ans += i;
            continue;
        }
        int cnt = 1;
        for (int j = 1; j < n; j++)
            if (!tag[j] && !(a[j] & i)) cnt++;
        if (cnt >= m){
            for (int j = 1; j < n; j++)
                if (!tag[j] && (a[j] & i))
                    tag[j] = true;
        }
        else ans += i;
    }
    printf("%lld\n",ans);
    return 0;
}

BZOJ4245 [ONTAK2015]OR-XOR 【貪心】