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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#101953#6355. 5zhouhuanyiWA 2ms13652kbC++112.1kb2023-05-01 21:55:232023-05-01 21:55:24

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-05-01 21:55:24]
  • 评测
  • 测评结果:WA
  • 用时:2ms
  • 内存:13652kb
  • [2023-05-01 21:55:23]
  • 提交

answer

#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cassert>
#define N 400000
#define M 20
#define inf 1e9
using namespace std;
int read()
{
    char c=0;
    int sum=0;
    while (c<'0'||c>'9') c=getchar();
    while ('0'<=c&&c<='9') sum=sum*10+c-'0',c=getchar();
    return sum;
}
struct reads
{
    int sz,p[M+1];
};
int n,S,cnt,cnt2,a[N+1],cnts[N+1];
long long ans;
reads c,dp[N+1],A[N+1],B[N+1],st[N+1];
reads operator + (reads a,reads b)
{
    int ps=0;
    for (int i=1;i<=b.sz;++i)
    {
	while (ps+1<=a.sz&&a.p[ps+1]<b.p[i])
	{
	    ++ps;
	    while (c.sz>=2&&a.p[ps]-c.p[c.sz-1]<=cnt2) c.sz--;
	    c.p[++c.sz]=a.p[ps];
	}
	while (c.sz>=2&&b.p[i]-c.p[c.sz-1]<=cnt2) c.sz--;
	c.p[++c.sz]=b.p[i];
    }
    for (int i=ps+1;i<=a.sz;++i)
    {
	while (c.sz>=2&&a.p[i]-c.p[c.sz-1]<=cnt2) c.sz--;
	c.p[++c.sz]=a.p[i];
    }
    return c;
}
reads operator + (reads a,int d)
{
    for (int i=1;i<=a.sz;++i) a.p[i]+=d;
    return a;
}
reads operator - (reads a,int d)
{
    for (int i=1;i<=a.sz;++i) a.p[i]-=d;
    return a;
}
int main()
{
    int d,sr,res;
    n=read(),S=read();
    for (int i=1;i<=n;++i)
    {
	a[i]=read()-1;
	if (a[i]==-1) cnt++;
	if (a[i]==0) cnt2++;
	if (a[i]>=1) cnts[a[i]]++;
    }
    res=cnt;
    for (int i=0;i<=cnt;++i) dp[i].p[dp[i].sz=1]=cnt-i;
    for (int i=1;i<=S;++i)
	if (cnts[i])
	{
	    res+=cnts[i]*i;
	    for (int j=0;j<i;++j)
	    {
		d=(res-j)/i;
		for (int k=0;k<=d;++k) st[k]=dp[k*i+j]-k;
		for (int k=0;k<=d;++k)
		{
		    if (!k||k/cnts[i]!=(k-1)/cnts[i]) A[k]=st[k];
		    else A[k]=A[k-1]+st[k];
		}
		for (int k=d;k>=0;--k)
		{
		    if (k==d||k/cnts[i]!=(k+1)/cnts[i]) B[k]=st[k];
		    else B[k]=B[k+1]+st[k];
		}
		for (int k=0;k<=d;++k)
		{
		    dp[k*i+j]=A[k];
		    if (k>=cnts[i]) dp[k*i+j]=dp[k*i+j]+B[k-cnts[i]];
		    dp[k*i+j]=dp[k*i+j]+k;
		}
	    }
	}
    for (int i=0;i<=res;++i)
    {
	sr=-inf;
	for (int j=1;j<=dp[i].sz;++j) ans+=dp[i].p[j]+cnt2-max(dp[i].p[j],sr+1)+1,sr=dp[i].p[j]+cnt2;
    }
    printf("%lld\n",ans);
    return 0;
}

詳細信息

Test #1:

score: 0
Wrong Answer
time: 2ms
memory: 13652kb

input:

7 9
0 0 0 1 1 2 5

output:

31

result:

wrong answer 1st numbers differ - expected: '42', found: '31'