NOIP 模拟 $30; m 毛三琛$

题解 (by;zjvarphi)

二分答案，考虑二分背包中的最大值是多少。

枚举 (p) 的值，在当前最优答案不优时，直接跳掉。

随机化一下 (p)，这样复杂度会有保证。

Code

#include<bits/stdc++.h>
#define ri register signed
#define p(i) ++i
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++
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            ri f=1;x=0;register char ch=gc();
            while(!isdigit(ch)) {if (ch=='-') f=0;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;}
    static const int N=1e4+7;
    int a[N],tmp[N],p[N],ans,n,P,k;
    inline int check(int mid) {
        ri cnt(0),nw(0);
        for (ri i(1);i<=n;p(i)) {
            if (tmp[i]>mid) return 0;
            if (nw+tmp[i]>mid) p(cnt),nw=0;
            nw+=tmp[i];
        }
        return cnt<k;
    }
    inline int MD(int x) {return x>=P?x-P:x;}
    inline int main() {
        //FI=freopen("nanfeng.in","r",stdin);
        //FO=freopen("nanfeng.out","w",stdout);
        srand(time(0)*clock()^time(0)*clock());
        cin >> n >> P >> k;
        for (ri i(1);i<=n;p(i)) cin >> a[i];
        for (ri i(1);i<=P;p(i)) p[i]=i-1;
        std::random_shuffle(p+1,p+P+1);
        ans=10000*n;
        for (ri i(1);i<=P;p(i)) {
            ri cp=p[i];
            for (ri j(1);j<=n;p(j)) tmp[j]=MD(a[j]+cp);
            if (!check(ans)) continue;  
            ri l(0),r(ans),res;
            while(l<=r) {
                int mid(l+r>>1);
                if (check(mid)) r=mid-1,res=mid;
                else l=mid+1;
            }
            ans=res;
        }
        printf("%d
",ans);
        return 0;
    }
}
int main() {return nanfeng::main();}