洛谷$P5444 [APIO2019]$奇怪装置 数论

正解:数论

解题报告:

传送门$QwQ$

我好像当初考的时候这题爆零了,,,部分分都没想到,,,我真的好菜$kk$

考虑如果在$t_1,t_2$两个时刻有$x_1=x_2,y_1=y_2$是什么情况$QwQ$?

那就有$egin{cases}t_1+[frac{t_1}{B}]equiv t_2+[frac{t_2}{B}](mod A)\t_1equiv t_2(mod B)end{cases}$.

不妨设$t_2=t_1+Bcdot t$,代入得$t_1+[frac{t_1}{B}]equiv t_1+Bcdot k+[frac{t_1+Bcdot k}{B}](mod A)$,即$k(B+1)equiv 0(mod A)$

解得$frac{A}{gcd(A,B+1)}|k$.即将$mod B$相等的提出来,发现每$frac{A}{gcd(A,B+1)}$一循环.又因为$mod B$的结果有$B$个,所以总的循环节长度为$len=frac{Acdot B}{gcd(A,B+1)}$.

所以把$l,r$取模后变成若干条线段,然后现在询问$[0,len)$覆盖了多少个点.昂这个不差分下就行了嘛,$over$

 

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define fi first
#define sc second
#define ll long long
#define gc getchar()
#define mp make_pair
#define P pair<ll,ll>
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ll i=x;i<=y;++i)
#define my(i,x,y) for(ll i=x;i>=y;--i)

const int N=1e6+10;
const ll inf=1e18;
ll n,A,B,len,l[N],r[N],sum,dat,lst,as;
multiset<P>S;

il ll read()
{
    rc ch=gc;ll x=0;rb y=1;
    while(ch!='-' && (ch>'9' || ch<'0'))ch=gc;
    if(ch=='-')ch=gc,y=0;
    while(ch>='0' && ch<='9')x=(x<<1)+(x<<3)+(ch^'0'),ch=gc;
    return y?x:-x;
}
ll gcd(ll x,ll y){return y?gcd(y,x%y):x;}
il void ad(ll x,ll y){S.insert(mp(x,-1));S.insert(mp(y,1));}

int main()
{
    //freopen("5444.in","r",stdin);freopen("5444.out","w",stdout);
    n=read();A=read();B=read();len=A/gcd(A,B+1)*B;rp(i,1,n){l[i]=read(),r[i]=read();sum+=r[i]-l[i]+1;}
    if(A/gcd(A,B+1)>inf/B)return printf("%lld
",sum),0;
    rp(i,1,n)
    {
        if(r[i]-l[i]>=len)return printf("%lld
",len),0;;//printf("l=%lld r=%lld len=%lld
",r[i],l[i],len);
        if(l[i]/len!=r[i]/len)ad(l[i]%len,len-1),ad(0,r[i]%len);else ad(l[i]%len,r[i]%len);
    }
    multiset<P>::iterator it=S.begin();
    while(it!=S.end()){P tmp=*it;if(!dat)lst=tmp.fi;dat-=tmp.sc;if(!dat)as+=tmp.fi-lst+1;++it;}
    printf("%lld
",as);
    return 0;
}