【Mail.Ru Cup 2018 Round 2 C】 Lucky Days

【链接】 我是链接,点我呀:)

【题意】

【题解】


题解的作者: manish_joshi

对于任意一个k

因为那条直线(关于x,y的方程可以看出一条直线)的斜率>=

所以肯定会经过第一象限。

因此总是能得到一个k1,k2>=0

连续的1的起点尽可能接近。

【代码】

#include <bits/stdc++.h>
#define LL long long
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
using namespace std;

const int N = 1e5;

LL la,ra,ta;
LL lb,rb,tb;

int main(){
    //freopen("D:\\rush.txt","r",stdin);
    scanf("%lld%lld%lld",&la,&ra,&ta);
    scanf("%lld%lld%lld",&lb,&rb,&tb);
    if (la>lb){
        swap(la,lb);
        swap(ra,rb);
        swap(ta,tb);
    }
    //la <= lb
    /*
        st1 = la + k1*t1
        st2 = lb + k2*t2
        dif = st2-st1
            = lb-la+k2*t2-k1*t1 = 0

        lb-la+k2*t2=k1*t1
        t1*k1=k2*t2+lb-la

        dir = lb-la+k*gcd(t1,t2)
        la-lb=k*gcd(t1,t2);
        k = (la-lb)/gcd(t1,t2);
    */
    LL temp1 = la-lb;
    LL temp2 = __gcd(ta,tb);
    if (temp1%temp2==0){
        LL ans = min(rb-lb+1,ra-la+1);
        printf("%lld\n",ans);
    }else{
        double temp = 1.0*temp1/temp2;

        LL len1 = ra-la+1,len2 = rb-lb+1;
        LL k = (LL) (ceil(temp));

        LL delta = lb-la+k*__gcd(ta,tb);

        LL aa = 1,bb = len1,cc = 1+delta,dd = delta+len2;
        LL ans = 0;
        ans = max(ans,min(bb,dd)-max(aa,cc)+1);

        k = (LL) (floor(temp));
        delta = lb-la+k*__gcd(ta,tb);
        aa = 1,bb = len1,cc = 1+delta,dd = delta+len2;
        ans = max(ans,min(bb,dd)-max(aa,cc)+1);

        printf("%lld\n",ans);
    }
    return 0;
}