Codeforces Round #287 (Div. 2) (question A, question B ),

Codeforces Round #287 (Div. 2) (question A, question B ),

PS: makeup. I forgot to send it in the past few days. At that time, SLB in the dormitory cried. He didn't squeeze in for the first 20 minutes. Finally, he couldn't move the computer to the fourth floor, sad. I made two questions and understood them later. However, due to poor learning skills, sad became a problem.

A. Amr and Musictime limit per test 1 secondmemory limit per test 256 megabytesinput standard inputoutput standard output

Amr is a young coder who likes music a lot. He always wanted to learn how to play music but he was busy coding so he got an idea.

Amr hasNInstruments, it takesAIDays to learnI-Th instrument. Being busy, Amr dedicatedKDays to learn how to play the maximum possible number of instruments.

Amr asked for your help to distribute his free days between instruments so that he can achieve his goal.

Input

The first line contains two numbersN,K(1 digit ≤ DigitNLimit ≤ limit 100, 0 limit ≤ limitKLimit ≤ limit 10 limit 000), the number of instruments and number of days respectively.

The second line containsNIntegersAI(1 digit ≤ DigitAILimit ≤ limit 100), representing number of days required to learnI-Th instrument.

Output

In the first line output one integerMRepresenting the maximum number of instruments Amr can learn.

In the second line outputMSpace-separated integers: the indices of instruments to be learned. You may output indices in any order.

If there are multiple optimal solutions output any. It is not necessary to use all days for studying.

Sample test (s) input

4 104 3 1 2

Output

41 2 3 4

Input

5 64 3 1 1 2

Output

31 3 4

Input

1 34

Output

0

Note

In the first test Amr can learn all 4 instruments.

In the second test other possible solutions are: {2, limit 3, limit 5} or {3, limit 4, limit 5 }.

In the third test Amr doesn't have enough time to learn the only presented instrument.

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <queue>using namespace std;int n,m;struct node{    int x,y;}a[11000];int cmp(struct node a,struct node b){    return a.x>b.x;}int b[11000];int main(){    int i;    while(~scanf("%d %d",&n,&m))    {        for(i=0;i<n;i++)        {            scanf("%d",&a[i].x);            a[i].y = i+1;        }        sort(a,a+n,cmp);        int t = 0;         memset(b,0,sizeof(b));        for(i=0;i<n;i++)        {            if(m>=a[i].x)            {                m = m-a[i].x;                b[t++] = a[i].y;            }            else                break;        }        sort(b,b+t);        printf("%d\n",t);        for(i=0;i<t;i++)        {            if(i==t-1)             printf("%d\n",b[i]);            else             printf("%d ",b[i]);        }    }    return 0;}

B. Amr and Pinstime limit per test 1 secondmemory limit per test 256 megabytesinput standard inputoutput standard output

Amr loves Geometry. One day he came up with a very interesting problem.

Amr has a circle of radiusRAnd center in point (X, Bytes,Y). He wants the circle center to be in new position (X', Region ',Y').

In one step Amr can put a pin to the border of the circle in a certain point, then rotate the circle around that pin by any angle and finally remove the pin.

Help Amr to achieve his goal in minimum number of steps.

Input

Input consists of 5 space-separated integersR,X,Y,X'Y'(1 hour ≤ hourRLimit ≤ limit 105, minimum-limit 105 limit ≤ limitX, Bytes,Y, Bytes,X', Region ',Y'Cycle ≤ cycle 105), circle radius, coordinates of original center of the circle and coordinates of destination center of the circle respectively.

Output

Output a single integer-minimum number of steps required to move the center of the circle to the destination point.

Sample test (s) input

2 0 0 0 4

Output

1

Input

1 1 1 4 4

Output

3

Input

4 5 6 5 6

Output

0

Note

In the first sample test the optimal way is to put a pin at point (0, limit 2) and rotate the circle by 180 degrees counter-clockwise (or clockwise, no matter ).

# Include <stdio. h> # include <string. h> # include <stdlib. h> # include <iostream> # include <algorithm> # include <set> # include <map> # include <queue> using namespace std; int main () {double r, x1, y1, x2, y2; while (~ Scanf ("% lf", & r, & x1, & y1, & x2, & y2) {double q = (double) sqrt (x1-x2) * (x1-x2) + (y1-y2) * (y1-y2); // the linear distance between the original center and the current center of the int cnt = 0; if (q = 0) {printf ("0 \ n");} else if (q <= 2 * r) {printf ("1 \ n ");} else {while (q> 2 * r) {q = q-2 * r; cnt ++;} printf ("% d \ n", cnt + 1 ); // Add 1 to indicate the situation in it. } Return 0 ;}