bzoj [Noi2002]Savage 扩展欧几里得
枚举m,n^2判断
对于野人i,j,(H[i]+x*S[i])%m==(H[j]+x*S[j])%m,且x<=O[i]&&x<=O[j],他们才有可能相遇
化简得:(S[i]-S[j])*x+y*m=C[j]-C[i],扩欧解x最小值,判断
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #define N 18 using namespace std; int H[N],S[N],O[N],n,m,X,Y; bool bo; int exgcd(int a,int b,int &x,int &y){ if(b==0){ x=1;y=0; return a; } int gcd=exgcd(b,a%b,x,y); int t=x; x=y; y=t-(a/b)*x; return gcd; } bool judge(int x,int y,int mm){ //printf("%d %d %d\n",x,y,mm); if(S[x]<S[y]) swap(x,y); int a=S[x]-S[y],b=mm,c=H[y]-H[x]; int d=exgcd(a,b,X,Y); if(c%d!=0) return 0; X*=c/d; int bd=b/d; X=((X%bd)+bd)%bd; if(X<=O[x]&&X<=O[y]) return 1; return 0; } int main(){ scanf("%d",&n); for(int i=1;i<=n;i++){ scanf("%d%d%d",&H[i],&S[i],&O[i]); m=max(m,H[i]); H[i]--; } for(;;m++){ bo=0; for(int i=1;i<=n;i++){ for(int j=i+1;j<=n;j++) if(judge(i,j,m)){bo=1;break;} if(bo==1)break; }if(bo==0)break; } printf("%d\n",m); return 0; }
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