题解 \(by\;zj\varphi\)
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\n",ans);
return 0;
}
}
int main() {return nanfeng::main();}