Educational Codeforces Round 52 (Rated for Div. 2)

A.

A. Vasya and Chocolate

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

There is a special offer in Vasya's favourite supermarket: if the customer buys $\mathrm{aa chocolate\; bars,\; he\; or\; she\; may\; take bb additional\; bars\; for\; free.\; This\; special\; offer\; can\; be\; used\; any\; number\; of\; times.}$

Vasya currently has $\mathrm{ss roubles,\; and\; he\; wants\; to\; get\; as\; many\; chocolate\; bars\; for\; free.\; Each\; chocolate\; bar\; costs cc roubles.\; Help\; Vasya\; to\; calculate\; the\; maximum\; possible\; number\; of\; chocolate\; bars\; he\; can\; get!}$

Input

The first line contains one integer $\mathrm{tt (1\le t\le 1001\le t\le 100)\; —\; the\; number\; of\; testcases.}$

Each of the next $\mathrm{tt lines\; contains\; four\; integers s,a,b,c (1\le s,a,b,c\le 109)s,a,b,c (1\le s,a,b,c\le 109) —\; the\; number\; of\; roubles\; Vasya\; has,\; the\; number\; of\; chocolate\; bars\; you\; have\; to\; buy\; to\; use\; the\; special\; offer,\; the\; number\; of\; bars\; you\; get\; for\; free,\; and\; the\; cost\; of\; one\; bar,\; respectively.}$

Output

Print $\mathrm{tt lines. ii-th\; line\; should\; contain\; the\; maximum\; possible\; number\; of\; chocolate\; bars\; Vasya\; can\; get\; in ii-th\; test.}$

Example

input

Copy

2
10 3 1 1
1000000000 1 1000000000 1

output

Copy

13
1000000001000000000

Note

In the first test of the example Vasya can buy $\mathrm{99 bars,\; get 33 for\; free,\; buy\; another\; bar,\; and\; so\; he\; will\; get 1313 bars.}$

In the second test Vasya buys $\mathrm{10000000001000000000 bars\; and\; gets 10000000000000000001000000000000000000 for\; free.\; So\; he\; has 10000000010000000001000000001000000000 bars.}$

#include <iostream>
#include <cstdio>

using namespace std;

int main()
{
    long long t,s,a,b,c;
    scanf("%lld",&t);
    while(t--){
        scanf("%lld %lld %lld %lld",&s,&a,&b,&c);
        long long temp=a*c;
        if(s<temp){
            printf("%lld
",s/c);
        }else{
            long long cou=(s/temp)*(a+b)+(s%temp)/c;
            printf("%lld
",cou);
        }
    }
    return 0;
}

B.

B. Vasya and Isolated Vertices

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has got an undirected graph consisting of $\mathrm{nn vertices\; and mm edges.\; This\; graph\; doesn\text{'}t\; contain\; any\; self-loops\; or\; multiple\; edges.\; Self-loop\; is\; an\; edge\; connecting\; a\; vertex\; to\; itself.\; Multiple\; edges\; are\; a\; pair\; of\; edges\; such\; that\; they\; connect\; the\; same\; pair\; of\; vertices.\; Since\; the\; graph\; is\; undirected,\; the\; pair\; of\; edges (1,2)(1,2) and (2,1)(2,1) is\; considered\; to\; be\; multiple\; edges.\; Isolated\; vertex\; of\; the\; graph\; is\; a\; vertex\; such\; that\; there\; is\; no\; edge\; connecting\; this\; vertex\; to\; any\; other\; vertex.}$

Vasya wants to know the minimum and maximum possible number of isolated vertices in an undirected graph consisting of $\mathrm{nn vertices\; and mm edges.}$

Input

The only line contains two integers $\mathrm{nn and m (1\le n\le 105,0\le m\le n(n-1)2)m (1\le n\le 105,0\le m\le n(n-1)2).}$

It is guaranteed that there exists a graph without any self-loops or multiple edges with such number of vertices and edges.

Output

In the only line print two numbers $\mathrm{minmin and maxmax —\; the\; minimum\; and\; maximum\; number\; of\; isolated\; vertices,\; respectively.}$

Examples

input

Copy

4 2

output

Copy

0 1

input

Copy

3 1

output

Copy

1 1

Note

In the first example it is possible to construct a graph with $\mathrm{00 isolated\; vertices:\; for\; example,\; it\; should\; contain\; edges (1,2)(1,2) and (3,4)(3,4).\; To\; get\; one\; isolated\; vertex,\; we\; may\; construct\; a\; graph\; with\; edges (1,2)(1,2) and (1,3)(1,3).}$

In the second example the graph will always contain exactly one isolated vertex.

wrong到结束......

还是不对

#include <iostream>
#include <cstdio>

using namespace std;

typedef long long ll;

ll cc[100005];

int main()
{
    ll n,m;
    for(ll i=1;i<=100004;i++){
        cc[i]=i*(i-1)/2;
    }
    while(~scanf("%lld %lld",&n,&m)){
        ll minn,maxx;
        if(n==1){
            printf("1 1
");
            continue;
        }
        if(n==0){
            printf("0 0
");
            continue;
        }
        if(m==0){
            printf("%lld %lld
",n,n);
            continue;
        }
        if(m*2>=n){
            minn=0;
            if(m>cc[n-1]){
                maxx=0;
            }else{
                for(int i=1;i<=99999;i++){
                    if(m>cc[i]&&m<=cc[i+1]){
                        maxx=n-(i+1);
                        break;
                    }
                }

            }
        }else{
            minn=n-2*m;
            maxx=n-m-1;
        }
        printf("%lld %lld
",minn,maxx);
    }
    return 0;
}