BZOJ 1407 Savage(拓展欧几里得)

这题的时间复杂度真玄学。。。 O(m*n^2)。1e8也能过啊。。。

首先题目保证m<=1e6. 这启发我们枚举或者二分答案？

但是答案不满足单调性，考虑从小到大枚举m。

对于每一个m，枚举两个野人在有生之年能否住在一起。可以推出一个同余方程，用扩欧可以求出最小整数解x，或者没有解。

如果x<=life[i]&&x<=life[j]那么当然不满足条件。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi 3.1415926535

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

int C[N], P[N], L[N], n;

int extend_gcd(int a, int b, int &x, int &y){

    if (a==&&b==) return -;

    if (b==){x=; y=; return a;}

    int d=extend_gcd(b,a%b,y,x);

    y-=a/b*x;

    return d;

}

bool check(int ans){

    int d, x, y;

    FOR(i,,n) FOR(j,i+,n) {

        d=extend_gcd(P[i]-P[j],-ans,x,y);

        if ((C[j]-C[i])%d) continue;

        x*=((C[j]-C[i])/d);

        int k=abs(-ans/d);

        x=(x%k+k)%k;

        if (x<=L[i]&&x<=L[j]) return false;

    }

    return true;

}

int main ()

{

    int ans=;

    scanf("%d",&n);

    FOR(i,,n) scanf("%d%d%d",C+i,P+i,L+i), ans=max(ans,C[i]);

    for (;;++ans) if (check(ans)) break;

    printf("%d\n",ans);

    return ;

}