标签：names int name nbsp max iostream clu 时间复杂度 splay
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-9 # define MOD 100000007 # define INF 1000000000 # 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<<1,l,mid # define rch p<<1|1,mid+1,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=0, flag=0; char ch; if((ch=getchar())==‘-‘) flag=1; else if(ch>=‘0‘&&ch<=‘9‘) res=ch-‘0‘; while((ch=getchar())>=‘0‘&&ch<=‘9‘) res=res*10+(ch-‘0‘); return flag?-res:res; } void Out(int a) { if(a<0) {putchar(‘-‘); a=-a;} if(a>=10) Out(a/10); putchar(a%10+‘0‘); } const int N=20; //Code begin... int C[N], P[N], L[N], n; int extend_gcd(int a, int b, int &x, int &y){ if (a==0&&b==0) return -1; if (b==0){x=1; y=0; 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,1,n) FOR(j,i+1,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=0; scanf("%d",&n); FOR(i,1,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 0; }
标签：names int name nbsp max iostream clu 时间复杂度 splay
原文地址：http://www.cnblogs.com/lishiyao/p/6568202.html
