BZOJ3721: PA2014 Final Bazarek

要求和为奇数，就是选奇数个奇数，任意个偶数
将n个数按奇偶分类，f[i],g[i]表示取i个数和最大，取了f[i]个奇数，g[i]个偶数，显然都是各自最大的f[i],g[i]个
转移的时候枚举取出的是奇数还是偶数，从i-2,i-1转移到i

code：

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long longusing namespace std;inline void read(int &x){    char c; while(!((c=getchar())>='0'&&c<='9'));    x=c-'0';    while((c=getchar())>='0'&&c<='9') (x*=10)+=c-'0';}const int maxn = 1100000;int n,m;int a[maxn],b[maxn],an,bn;ll ans[maxn];int f[maxn],g[maxn];int main(){    read(n);    for(int i=1;i<=n;i++)    {        int x; read(x);        if(x&1) a[++an]=x;        else b[++bn]=x;    }    sort(a+1,a+an+1);    sort(b+1,b+bn+1);    memset(ans,-1,sizeof ans);    if(an) ans[1]=a[an],f[1]=1,g[1]=0;    for(int i=2;i<=n;i++)    {        if(f[i-2]+2<=an&&ans[i-2]!=-1)        {            int p=an-f[i-2];            if(ans[i]<ans[i-2]+a[p]+a[p-1]) ans[i]=ans[i-2]+a[p]+a[p-1],f[i]=f[i-2]+2,g[i]=g[i-2];        }        if(g[i-1]+1<=bn&&ans[i-1]!=-1)        {            int p=bn-g[i-1];            if(ans[i]<ans[i-1]+b[p]) ans[i]=ans[i-1]+b[p],f[i]=f[i-1],g[i]=g[i-1]+1;        }    }    read(m);    while(m--)    {        int x; read(x);        printf("%lld\n",ans[x]);    }    return 0;}