Modified LCS

Input

Output

Sample Input

3
5 3 4 15 3 1
10 2 2 7 3 3
100 1 1 100 1 2

Sample Output

4
3
50
超时代码，因为K很大

 /*****************
 f1+(k1-1)*d1=f2+(k2-1)*d2
 =>  (k1-1)*d1+(k2-1)*(-d2)=f2-f1;
 =>  d1*x+y*(-d2)=f2-f1   d1=>[0,n1-1]  d2=>[0,n2-1]
 求在给定范围内解的个数
 *********************/
 #include"iostream"
 #include"cstdio"
 #include"cstring"
 #include"cmath"
 #include"algorithm"
 using namespace std;
 typedef long long ll;
 void exgcd(ll n,ll m,ll &x,ll &y,ll &d)
 {
     if(!m){
         x=;y=;d=n;
     }
     else{
         exgcd(m,n%m,y,x,d);
         y-=(n/m)*x;
     }
 }
 int main()
 {
     ll f1,n1,d1,f2,n2,d2,i,j,t;
     cin>>t;
     while(t--)
     {
         ll x,y,tmp,d;
         //cin>>n1>>f1>>d1>>n2>>f2>>d2;
         scanf("%lld%lld%lld%lld%lld%lld",&n1,&f1,&d1,&n2,&f2,&d2);
         exgcd(d1,-d2,x,y,d);
         f2-=f1;
         if(f2%d){
             //cout<<"0"<<endl;
             puts("");
             continue;
         }
         x=x*(f2/d);
         y=y*(f2/d);
         //接下来  逼近[0,n1-1],[0,n2-1]
         ll t1=d2/(abs(d));
         ll t2=d1/(abs(d));
         if(x<||y<)
         {
             int i=;
             while()
             {
                 if(x+t1*i>=&&y+t2*i>=)
                     break;
                 i++;
             }
             x=x+t1*i;
             y=y+t2*i;
         }
         else
         {
             int i=;
             while()
             {
                 if(x-t1*i<||y-t2*i<)
                     break;
                 i++;
             }
             x=x-t1*(i-);
             y=y-t2*(i-);
         }
         if(x>(n1-)||y>(n2-))
         {
             //cout<<"0"<<endl;
             puts("");
             continue;
         }
         int ans=;
         while()
         {
             if(x+t1*ans>(n1-)||y+t2*ans>(n2-))
                 break;
             ans++;
         }
         cout<<ans<<endl;
         //printf("%lld\n",ans);
     }
     return ;
 }

AC代码

/*****************
f1+(k1-1)*d1=f2+(k2-1)*d2
=>  (k1-1)*d1+(k2-1)*(-d2)=f2-f1;
=>  d1*x+y*(-d2)=f2-f1   d1=>[0,n1-1]  d2=>[0,n2-1]
求在给定范围内解的个数
*********************/
#include"iostream"
#include"cstdio"
#include"cstring"
#include"cmath"
#include"algorithm"
using namespace std;
typedef long long ll;
void exgcd(ll n,ll m,ll &x,ll &y,ll &d)
{
    if(!m){
        x=;y=;d=n;
    }
    else{
        exgcd(m,n%m,y,x,d);
        y-=(n/m)*x;
    }
}
int main()
{
    ll f1,n1,d1,f2,n2,d2,i,j,t;
    cin>>t;
    while(t--)
    {
        ll x,y,tmp,d;
        //cin>>n1>>f1>>d1>>n2>>f2>>d2;
        scanf("%lld%lld%lld%lld%lld%lld",&n1,&f1,&d1,&n2,&f2,&d2);
        exgcd(d1,-d2,x,y,d);
        f2-=f1;
        if(f2%d){
            //cout<<"0"<<endl;
            puts("");
            continue;
        }
        x=x*(f2/d);
        y=y*(f2/d);
        //接下来  逼近[0,n1-1],[0,n2-1]
        ll t1=d2/(abs(d));
        ll t2=d1/(abs(d));
        if(x<||y<)
        {
            int i=;
            while()
            {
                if(x+t1*i>=&&y+t2*i>=)
                    break;
                i++;
            }
            x=x+t1*i;
            y=y+t2*i;
        }
        else
        {
            int i=;
            while()
            {
                if(x-t1*i<||y-t2*i<)
                    break;
                i++;
            }
            x=x-t1*(i-);
            y=y-t2*(i-);
        }
        if(x>(n1-)||y>(n2-))
        {
            //cout<<"0"<<endl;
            puts("");
            continue;
        }
        int ans=;
    /*    while(1)
        {
            if(x+t1*ans>(n1-1)||y+t2*ans>(n2-1))
                break;
            ans++;
        }
    */
    //  这样避免超时
        ll a,b;
        a=(n1--x)/t1;
        b=(n2--y)/t2;
        cout<<min(a,b)+<<endl;
    }
    return ;
}