Educational Codeforces Round 32 E. Maximum Subsequence

E. Maximum Subsequence

题意： n 个数，选出其中  k 个数，使得他们的和对 m 取模后最大。 输出这个最大值。

tags：注意到 n 很小， 所以折半枚举。

//  E
#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 800005;

ll  n, m, s1[N], s2[N], a[N];
int main()
{
    scanf("%lld%lld", &n, &m);
    rep(i,1,n) scanf("%lld", &a[i]);
    int n1 = n/2, n2 = n-n1;
    int cnt1 = (1<<n1)-1, cnt2 = (1<<n2)-1;
    for(int i=0; i<=cnt1; ++i)
    {
        ll num = 0;
        for(int j=0; j<n1; ++j)
            if((i>>j)&1) num += a[j+1];
        s1[i] = num%m;

    }
    for(int i=0; i<=cnt2; ++i)
    {
        ll num = 0;
        for(int j=0; j<n2; ++j)
            if((i>>j)&1) num += a[n1+j+1];
        s2[i] = num%m;
    }
    sort(s2, s2+cnt2+1);
    sort(s1, s1+cnt1+1);
    cnt1 = unique(s1, s1+cnt1+1) - s1;
    cnt2 = unique(s2, s2+cnt2+1) - s2;
    --cnt1, --cnt2;
    ll  ans = 0;
    for(int i=0; i<=cnt1; ++i)
    {
        ll num = m-s1[i];
        int pos = lower_bound(s2, s2+cnt2+1, num) - s2;
        if(pos==cnt2+1 || pos>0)
            ans = max(ans, s1[i]+s2[pos-1]);
    }
    printf("%lld\n", ans);

    return 0;
}

原文地址：https://www.cnblogs.com/sbfhy/p/7814640.html