[Codeforces958C3][DP][树状数组]Encryption (hard)

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翻译

给你一个长度为n的序列，要求你把它分成K段
每段的价值为这段的总权值%P
要求总价值最小
n<=500000 K<=100 P<=100

题解

方程
$f[i][j]=min(f[k][j-1]+(sum[i]-sum[k])\mod P)$
设 $s[i]=sum[i]\mod P$
分 $s[i]&gt;s[k]$与 $s[i]&lt;s[k]$讨论
变为 $s[i]-s[k]$和 $s[i]-s[k]+P$
发现 $s[i]$的值域很小 可以对于每个s[i]，维护最小的 $f[i]-s[i]$
树状数组搞一搞就好了

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<ctime>
#include<map>
#define LL long long
#define mp(x,y) make_pair(x,y)
using namespace std;
inline int read()
{
	int f=1,x=0;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline void write(int x)
{
	if(x<0)putchar('-'),x=-x;
	if(x>9)write(x/10);
	putchar(x%10+'0');
}
inline void print(int x){write(x);printf(" ");}
int s1[2][110];
int s[510000],n,K,mod;
inline void _min(int &x,int y){x=x<y?x:y;}
inline void ad(int &x,int y){x+=y;if(x>mod)x-=mod;}
inline int lowbit(int x){return x&-x;}
inline void change(int x,int c,int op){for(;x<=mod;x+=lowbit(x))_min(s1[op][x],c);}
inline int findmin(int x,int op){int ret=2147483647;for(;x>=1;x-=lowbit(x))_min(ret,s1[op][x]);return ret;}
int f[110][510000];
int val[110];
int main()
{
	n=read();K=read();mod=read();
	memset(f,63,sizeof(f));
	for(int i=1;i<=n;i++)
	{
		s[i]=(read()%mod+s[i-1])%mod;
		f[1][i]=s[i];
	}
	for(register int k=2;k<=K;k++)
	{
		memset(s1,63,sizeof(s1));
		memset(val,63,sizeof(val));
		val[s[k-1]]=f[k-1][k-1]-s[k-1];
		change(s[k-1]+1,val[s[k-1]],0);
		change(mod-s[k-1],val[s[k-1]],1);
		for(register int i=k;i<=n;i++)
		{
			int g1=findmin(s[i]+1,0),g2=findmin(mod-s[i],1);
			_min(f[k][i],min(s[i]+g1,s[i]+mod+g2));
			if(val[s[i]]>f[k-1][i]-s[i])
			{
				val[s[i]]=f[k-1][i]-s[i];
				change(s[i]+1,val[s[i]],0);
				change(mod-s[i],val[s[i]],1);
			}
		}
	}
	printf("%d\n",f[K][n]);
	return 0;
}