Could someone tell me how to input double fast in C++ ? scanf is not fast enough

Edited by author 25.11.2014 08:55

10+1 + 5+1 + 2 = 19.
And there is no other way

You can use DFS find the tips as this.
[code deleted]

Edited by moderator 19.11.2019 22:57

Hi,

I had wrong answer on test 9. Then I found an error and got the problem accepted. My test may be useful:

1000 1000
1 1
2 1
1 2
50 50
1
10 10
11 10
10 11


69.392

0.343 sec.
Just to let you know. ;)

WTF

#include <stdio.h>
#include <math.h>
#include <iostream>
#include <stdlib.h>
using namespace std;
int main()
{
    long int range, total = 0;
    cin >> range;
    if ( range <= 0 && abs(range) < 10000 )
    {
        for ( int x = 1; x >= range; --x )
            total += x;
        cout << total << endl;
    }
    else if ( range > 0 && abs(range) < 10000 )
    {
        for ( int x = 1; x <= range; ++x )
            total += x;
        cout << total << endl;
    }



    return 0;
}

#include <math.h>
#include <iostream>
using namespace std;

int main()
{
    double *a = new double[131073];
    double temp;
    int i = 0;
    while (cin >> temp) a[i++] = sqrt(temp);
    while (i > 0)
    {
        printf("%.4f\n", a[--i]);
    }
    delete[] a;
    return 0;
}

What could test 43 be?
If the spell distroys 0 coins, is the minimum amount of spells also 0?
Or is it 1?

Can there be identical threads in the test?
Like:
4 4
1 2
1 2
2 3
3 4
3
1 2 3
And if they can occur is the answer 2 2 3 or 1 2 3?

I have got AC, but I have found test for which my AC program gives wrong answer. I sent it to timus_support@acm.timus.ru.

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int a, b, c[10000], n, p, sum = 0, i, k, s, m;
    scanf("%i %i", &a, &b);

    for(i = 1;i <= b ; i++)
    {
        scanf("%i", &c[i]);

        sum += c[i - 1];
    }
    k = sum - a * (b - 1);
    if(k <= 0) {
       m = c[b] - a;
       if(m <= 0)
        printf("0\n");
       else
        printf("%i\n", m);
    }
    else {
        n = (k + c [b]) - a;
        if(n <= 0)
            printf("0\n");
        else
            printf("%i\n", n);
    }
}

Edited by author 20.11.2014 20:03

n = 8, k = 3
one computer already has got prog installed, so you start copying from that one.
you copied to any one (to exactly one as problem requires), 1 hour passed, k = 2 (2 cables left).
you copied to next(any) one, 2 hours passed, k = 1 (and 1 cable left), copied to next - 3 hours passed and no cables left !! at the moment there are 4 computers remaining without programm, so as i see its impossible to copy to all computers, becuase no cables left ...
how the hell output is 4 ?? and how you use them 3 cables to copy to all computers ...
Whats wroung in this ???
help me please, thanks !

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int n, a;
    cin >> n >> a;

    if (n == 1)
        cout << 0 << endl;
    else
    {
        int S = n - 1;
        //int S = n;
        for (int i = 1; i <= n; i++)
        {
            S = S - min(a, i);

            if (S <= 0)
            {
                cout << i << endl;
                break;
            }
        }
    }
}

can anybody give me some tests?

I've seen few people got AC in 0.015 seconds, but how? I used binary search and each iteration's left border started from the last found number, but not from the beginning of array and got 0.031 seconds.

var a:integer;
    b:integer;
    va:integer;
    vb:integer;
    ea:integer;
    eb:integer;
    ra:integer;
    rb:integer;
    boxa:integer;
    boxb:integer;
begin
    readln(a,b);
    readln(va,vb);

    boxb:=b-(va-a);
    rb:=b-boxb;
    readln(ea,eb);
    boxa:=va-(eb-vb);

    writeln(a-boxa,' ',b-boxb);

end.

/*
My program got WA on test 11!!!
What's wrong with my code?
Anyone can help me!!!
*/
#include <map>
#include <set>
#include <ctime>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <vector>
#include <string>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <iostream>
#include <algorithm>

#define FI            first
#define SE            second
#define LSON(x)            (x<<1)
#define RSON(x)            ((x<<1)|1)
#define Benefit(x,y)        x=max(x,y)
#define CS            const static
#define SORT_CMP(s,l,r,cmp)    sort(s+(l),s+(r)+1,cmp)
#define SORT(s,l,r)        sort(s+(l),s+(r)+1)
#define MP(x,y)            make_pair(x,y)
#define Randomize        srand( (unsigned int) time ( NULL ) )
#define INOUT(x,y)        freopen(x,"r",stdin),freopen(y,"w",stdout)

using namespace std;

typedef char name[20];

CS int MaxN = 100010 ;

struct word
{
    name x;
    int val,len;
    word()    {x[0] = 0; val = 0;}
};

struct Trie
{
    Trie * next[26];
    word * ans[10];
    Trie()
    {
        for(int i=0;i<26;i++)    next[i] = NULL ;
        for(int i=0;i<10;i++)    ans[i] = NULL ;
    }
} * Tree;

word p[MaxN];

name ask;

int n,m,asklen;

void Insert(Trie * x,int _y,int pos)
{
    int i;    word & y = p[_y];
    for(i=0;i<10;i++)
        if((x->ans[i]==NULL)||(x->ans[i]->val<y.val))    break;
    if(i<10)
    {
        for(int j=9;j>i;j--)
            x -> ans[j] = x->ans[j-1];
        x -> ans[i] = p + _y;
    }
    if(pos!=y.len)
    {
        int u = y.x[pos] - 'a';
        if(!x->next[u])    x -> next[u] = new Trie ;
        Insert(x->next[u],_y,pos+1);
    }
}

void Init()
{
    scanf("%d\n",&n);    Tree = new Trie ;
    for(int i=1;i<=n;i++)
    {
        scanf("%s %d\n",p[i].x,&p[i].val);    p[i].len=strlen(p[i].x);
        Insert(Tree,i,0);
    }
}

void Query(Trie * x,int pos)
{
    if(pos == asklen)
    {
        for(int i=0;i<10;i++)
            if(x->ans[i])    printf("%s\n",x->ans[i]->x);
            else        break;
        return ;
    }
    int u = ask[pos] - 'a';
    if(x->next[u])    Query(x->next[u],pos+1);
}

void Solve()
{
    scanf("%d\n",&m);
    while(m--)
    {
        scanf("%s\n",ask);    asklen = strlen(ask) ;
        Query(Tree,0);    if(m)    printf("\n");
    }
}

int main()
{
    Init();    Solve();
    return 0;
}

Edited by author 31.05.2012 14:40

Some hint:
For n = 4

1 : (1,2), (1,3), (1,4);
2 : (2,1), (2,3);
3 : (3,1), (3,2);
4 : (4,1);

1st have 3 friends, 2nd have 2 friends, 3d have 2 friends, 4fth have 1 friend, so ans is
4
1 2
1 3
1 4
2 3