Codeforces Round #287 (Div. 2)(A,B題)，

Codeforces Round #287 (Div. 2)(A,B題)，

PS：補發，這幾天忘記發了。當時被宿舍的渣比網給氣哭了，開始二十分鐘都沒有擠進去，最後沒法子搬著電腦黑燈瞎火的去了四樓，sad。做出兩道題來，後面的題看懂了，但是由於學藝不精，sad了。

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 has n instruments, it takes ai days to learn i-th instrument. Being busy, Amr dedicated k days 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 numbers n, k (1 ≤ n ≤ 100, 0 ≤ k ≤ 10 000), the number of instruments and number of days respectively.

The second line contains n integers ai (1 ≤ ai ≤ 100), representing number of days required to learn the i-th instrument.

Output

In the first line output one integer m representing the maximum number of instruments Amr can learn.

In the second line output m space-separated integers: the indices of instruments to be learnt. 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, 3, 5} or {3, 4, 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 radius r and center in point (x, y). He wants the circle center to be in new position (x', 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 integers r, x, y, x' y' (1 ≤ r ≤ 105,  - 105 ≤ x, y, x', y' ≤ 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, 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 %lf %lf %lf %lf",&r,&x1,&y1,&x2,&y2)) {        double q =(double)sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));//原圓心和現圓心的直線距離        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);//加上1是指在裡面的情況。        }    }    return 0;}