【JZOJ6384】珂学家

description

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analysis

• 注意配出来的饮料不可以再配成其他饮料，所以肯定有\(O(n^2)\)的枚举

• 而且可口度两两互不相同，搞得我以为这是神仙题

• 考虑把两个试剂\([l_1,r_1],[l_2,r_2]\)合并，\([l_1+l_2,r_1+r_2]\)里的每个点都有贡献

• 假设贡献只有\(1\)，拉出来就是\(1,2,3,..,k-1,k,k,k,...,k,k-1,...,3,2,1\)

• 注意到前面一段和后面一段是等差数列，中间全部相等，做一遍差分

• 就会是\(0,1,1,...,1,1,0,0,...,0,-1,...,-1,-1,-1,-1\)，注意最后面多出来一个\(-1\)

• 再做差分，变成\(0,1,0,...,0,0,-1,0,...,0,-1,...,0,0,0,0,1\)，最后面又多出来一个\(1\)

• 那么对于一次贡献，相当于打上两个加两个减标记，最后两次前缀和还原原数组

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code

#pragma GCC optimize("O3")
#pragma G++ optimize("O3")
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXN 5005
#define MAX 20000005
#define ha 998244353
#define ll long long
#define reg register ll
#define fo(i,a,b) for (reg i=a;i<=b;++i)
#define fd(i,a,b) for (reg i=a;i>=b;--i)

using namespace std;

ll v[MAXN],l[MAXN],r[MAXN];
ll a[MAX];
ll n,m,mx;

inline ll read()
{
	ll x=0,f=1;char ch=getchar();
	while (ch<'0' || '9'<ch){if (ch=='-')f=-1;ch=getchar();}
	while ('0'<=ch && ch<='9')x=x*10+ch-'0',ch=getchar();
	return x*f;
}
inline void swap(ll &x,ll &y){ll z=x;x=y,y=z;}
int main()
{
	//freopen("T1.in","r",stdin);
	freopen("a.in","r",stdin);
	freopen("a.out","w",stdout);
	n=read(),m=read();
	fo(i,1,n)v[i]=read(),l[i]=read(),mx=max(mx,r[i]=read());
	fo(i,1,n-1)fo(j,i+1,n)
	{
		ll lx=l[i],rx=r[i],ly=l[j],ry=r[j],k,kk,tmp=v[i]*v[j]%ha;
		(a[lx+ly]+=tmp)%=ha,(a[lx+ry+1]-=tmp)%=ha,(a[rx+ly+1]-=tmp)%=ha,(a[rx+ry+2]+=tmp)%=ha;
	}
	fo(i,1,2*mx+1)(a[i]+=a[i-1])%=ha;
	fo(i,1,2*mx+1)(a[i]+=a[i-1])%=ha;
	while (m--)printf("%lld\n",(a[read()]+ha)%ha);
	return 0;
}