NOIP 模拟 $90\; \rm 混乱邪恶$

题解 \(by\;zj\varphi\)

题意就是将数列分成两组，使得两组的和相等即可。

这个做法是乱搞，无法证明。

考虑贪心，钦定最大的数在其中一个集合，然后从大到小填数，只要当前数比剩下的值小，就填进去。

多枚举几个钦定的最大的数，即可提高正确性。

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define debug1(x) std::cerr << #x"=" << x << ' '
    #define debug2(x) std::cerr << #x"=" << x << std::endl
    #define Debug(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        } 
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    using ll=long long;
    static const int N=1e6+7;
    int a[N],p[N],c[N],n,m;
    ll hf,sum;
    inline int main() {
        FI=freopen("chaoticevil.in","r",stdin);
        FO=freopen("chaoticevil.out","w",stdout);
        cin >> n >> m;
        for (ri i(1);i<=n;pd(i)) cin >> a[i],sum+=a[p[i]=i];
        std::sort(p+1,p+n+1,[](int x,int y) {return a[x]>a[y];});
        hf=sum>>1;
        for (ri i(1);i<=n;pd(i))
            if (hf>=a[p[i]]) {
                hf-=a[p[i]];
                c[p[i]]=1;
            } else c[p[i]]=-1;
        if (!hf) {
            printf("NP-Hard solved\n");
            for (ri i(1);i<=n;pd(i)) printf("%d ",c[i]);
            printf("\n");
        } else {
            for (ri lg(2);lg<=5;pd(lg)) {
                hf=sum>>1;
                for (ri i(lg);i<n+lg;pd(i))
                    if (hf>=a[p[i%n]]) {
                        hf-=a[p[i%n]];
                        c[p[i%n]]=1;
                    } else c[p[i%n]]=-1;
                if (!hf) {
                    printf("NP-Hard solved\n");
                    for (ri i(1);i<=n;pd(i)) printf("%d ",c[i]);
                    return printf("\n"),0;
                }
            }
        }
        return 0;
    }
}
int main() {return nanfeng::main();}