cf 1063D Round 516 Candies for Children

D. Candies for Children

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

At the children's festival, children were dancing in a circle. When music stopped playing, the children were still standing in a circle. Then Lena remembered, that her parents gave her a candy box with exactly $\mathrm{kk candies\; "Wilky May".\; Lena\; is\; not\; a\; greedy\; person,\; so\; she\; decided\; to\; present\; all\; her\; candies\; to\; her\; friends\; in\; the\; circle.\; Lena\; knows,\; that\; some\; of\; her\; friends\; have\; a\; sweet\; tooth\; and\; others\; do\; not.\; Sweet\; tooth\; takes\; out\; of\; the\; box\; two\; candies,\; if\; the\; box\; has\; at\; least\; two\; candies,\; and\; otherwise\; takes\; one.\; The\; rest\; of\; Lena\text{'}s\; friends\; always\; take\; exactly\; one\; candy\; from\; the\; box.}$

Before starting to give candies, Lena step out of the circle, after that there were exactly $\mathrm{nn people\; remaining\; there.\; Lena\; numbered\; her\; friends\; in\; a\; clockwise\; order\; with\; positive\; integers\; starting\; with 11 in\; such\; a\; way\; that\; index 11 was\; assigned\; to\; her\; best\; friend\; Roma.}$

Initially, Lena gave the box to the friend with number $\mathrm{ll,\; after\; that\; each\; friend\; (starting\; from\; friend\; number ll)\; took\; candies\; from\; the\; box\; and\; passed\; the\; box\; to\; the\; next\; friend\; in\; clockwise\; order.\; The\; process\; ended\; with\; the\; friend\; number rr taking\; the\; last\; candy\; (or\; two,\; who\; knows)\; and\; the\; empty\; box.\; Please\; note\; that\; it\; is\; possible\; that\; some\; of\; Lena\text{'}s\; friends\; took\; candy\; from\; the\; box\; several\; times,\; that\; is,\; the\; box\; could\; have\; gone\; several\; full\; circles\; before\; becoming\; empty.}$

Lena does not know which of her friends have a sweet tooth, but she is interested in the maximum possible number of friends that can have a sweet tooth. If the situation could not happen, and Lena have been proved wrong in her observations, please tell her about this.

Input

The only line contains four integers $\mathrm{nn, ll, rr and kk (1\le n,k\le 10111\le n,k\le 1011, 1\le l,r\le n1\le l,r\le n) —\; the\; number\; of\; children\; in\; the\; circle,\; the\; number\; of\; friend,\; who\; was\; given\; a\; box\; with\; candies,\; the\; number\; of\; friend,\; who\; has\; taken\; last\; candy\; and\; the\; initial\; number\; of\; candies\; in\; the\; box\; respectively.}$

Output

Print exactly one integer — the maximum possible number of sweet tooth among the friends of Lena or "-1" (quotes for clarity), if Lena is wrong.

Examples

input

Copy

4 1 4 12

output

Copy

2

input

Copy

5 3 4 10

output

Copy

3

input

Copy

10 5 5 1

output

Copy

10

input

Copy

5 4 5 6

output

Copy

-1

Note

In the first example, any two friends can be sweet tooths, this way each person will receive the box with candies twice and the last person to take sweets will be the fourth friend.

In the second example, sweet tooths can be any three friends, except for the friend on the third position.

In the third example, only one friend will take candy, but he can still be a sweet tooth, but just not being able to take two candies. All other friends in the circle can be sweet tooths as well, they just will not be able to take a candy even once.

In the fourth example, Lena is wrong and this situation couldn't happen.

 

题目大意： 告诉你有n个人在圈里，共计有k个糖果，从起点出发到终点（可能经过好几圈），每个人只拿一个糖果或者两个糖果每次，爱吃甜食的小朋友可以拿两个一次（但是如果糖果不够了，也只能拿一个），问最多有多少个小朋友爱吃甜食。

n,k<=10^11

思路： 

n和k都比较大，所以我们要分类讨论。

考虑到经过t+1次的那部分有x个人，经过t次的那部分有y个人。x=(r-l+n)%n+1;  x里面有a个吃甜食，y里面有b个吃甜食

如果n比较小，那么我们就暴力枚举a和b的值。

如果n比较大，此时k/n比较小，也就是圈数比较少，我们可以暴力枚举t，然后解二元一次方程组。 

注意： 最后一个小朋友吃几个糖果要分类讨论（因为最后一个小朋友有可能喜欢吃甜食）。

#include<bits/stdc++.h>
using namespace std; 

#define F(i,a,b) for(int i=a;i<=b;i++) 
#define D(i,a,b) for(int i=a;i>=b;i--) 
#define ms(i,a)  memset(a,i,sizeof(a)) 
#define LL       long long 
template<class T>void read(T &x){
    x=0; char c=getchar(); 
    while (c<'0' || c>'9') c=getchar(); 
    while (c>='0' && c<='9') x=(x<<3)+(x<<1)+(c^48),c=getchar();
}
template<class T>void write(T x){
    if(x<0) x=-x,putchar('-'); 
    if(x>9) write(x/10); 
    putchar(48+x%10);  
}
template<class T>void chkmax(T &x,T y){ x=x>y? x:y; }  
LL n,k,l,r,x,t,s,p,ans,y;  
LL const inf=1LL<<50;  
LL calc(LL p,LL s,int a,int b){
    if(a==0 || b==1){
        if(k-s-x>=0 && (k-s-x)%(n+p)==0)
            return p;  
        else return -1;  
    }else {
        if(k-s+1-x>=0 && (k-s+1-x)%(n+p)==0)
            return p;  
        else return -1;  
    }
}

LL exgcd(LL &x,LL &y,LL a,LL b){
    if(!b){
        x=1;y=0; return a; 
    }
    LL t=exgcd(y,x,b,a%b);  
    y-=a/b*x; 
    return t;  
}
void solve(LL t1,LL t2,LL c,LL add){
    if(t2==0){
        if(2*x>=k && k>=x) chkmax(ans,min(x,k-x+1)+y); 
        return;
    }
    LL a,b;
    LL g=exgcd(a,b,t1,t2); 
    a=a*c; b=b*c; 
    LL l=-inf,r=inf;  
    a-=add; 
    if (a>=0) r=min(r,a/t2); else  r=min(r,(a-t2+1)/t2); a+=add;  
    if(a-x>=0) l=max(l,(a-x+t2-1)/t2); else l=max(l,(a-x)/t2);  
    if(y-b>=0) r=min(r,(y-b)/t1); else r=min(r,(y-b-t1+1)/t1); 
    if(-b>=0) l=max(l,(-b+t1-1)/t1); else l=max(l,-b/t1);  
    if(l<=r){
        chkmax(ans,a+b+l); 
        chkmax(ans,a+b+r);  
    }
}
int main(){
    read(n); 
    read(l); 
    read(r); 
    read(k); 
    x=(r-l+n)% n+1;  
    y=n-x;  
    ans=-1;  
    if(k/n>=5000000){
        F(p,0,n) F(s,max(0LL,x-(n-p)),min((int)x,p)){
            if(s<x) chkmax(ans,calc(p,s,0,0));  
            if(s)chkmax(ans,calc(p,s,1,0));  
            if(s)chkmax(ans,calc(p,s,1,1));  
        }
    }else {
        F(t,0,k/n){
            solve(t+1,t,k-t*x-x-t*y,0);  
            solve(t+1,t,k-t*x-x-t*y+1,1);  
        }
    }
    write(ans); 
    return 0; 
}