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

int solve(int n, string k){
    int ans=k.size();

    if(n>=1){
        return n*solve(n-ans);

    }
    else{
        return 1;


    }


}



int main()
{
   int  t,n;
    string k;
    cin>>n;
    cin>>k;
    int result=solve(n,k);
    cout<<result<<endl;


    return 0;
}

Please, give me test

This is Test 4

11
1 2 2 1 1 1 2 2 1 2 1
11
1 1 3
1 4 1
2 8 1
10 11 1
1 3 1
2 3 1
3 7 2
4 6 2
5 9 2
11 11 2
10 11 2

Answer is 01111010101

Edited by author 08.02.2013 04:06

Please, tell me how to solve this problem...
Not solution but an idea or a hint...
Thanks.

please give me some tests.

Those who are getting WA4 when using C#, please use the following format to print out the output. The type of variable sum is double and individual numbers read in the second line are also converted to double first before adding them together.

Console.WriteLine("{0:F6}", sum / cases);

Hello women and men! I haven't been able to solve this problem for a long time.
Please, help me that.

Does anyone see what goes wrong with my program for the input of test #9? or test inputs maybe?


EDIT: Test Case found!

Edited by author 01.06.2012 22:08

strips can go up

0 30
-25.9807621135 -15
25.9807621135 -15

3.711537444785714 10.714285714285715
-11.134612334357143 -2.142857142857144
7.423074889571431 -8.571428571428571
18.557687223928568 23.57142857142857
-29.69229955828571 4.285714285714285
11.134612334357145 -27.857142857142854

deleted

Edited by moderator 23.07.2022 20:36

Почему в случае 1 10 ответ 9 если 0 тоже число в котором не идут два подряд ноля?

Просто 2 раза получил ВА, поменял на 9 и АС.

Calculate matrix with modulo 10^9 + 7

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

int const N = 123456;
#define pi acos(-1.0)
typedef long long ll;

int ar[N],xar[N],uses[N];

struct points {
     double x, y;
     int id;
     points() {}
     points(double x, double y) : x(x), y(y) {}
 } ;

double ang(const points &p){
    double res = atan2(p.y, p.x);
    if(res < 0) res += 2.0 * pi;
    return res;
}

struct cmp{
    inline bool operator () (const points &p1, const points &p2){
        double ang1 = ang(p1)*(180/pi), ang2 = ang(p2)*(180/pi);
        if(fabs(ang1 - ang2) < 1e-9){
            ll d1 = (ll)p1.x * (ll)p1.x + (ll)p1.y * (ll)p1.y;
            ll d2 = (ll)p2.x * (ll)p2.x + (ll)p2.y * (ll)p2.y;

            return d1 < d2;
        }
        return ang1 < ang2;
    }
};

points pt[N];

int main() {
    int n;
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> pt[i].x >> pt[i].y;
        pt[i].id = i+1;
    }
    sort(pt, pt+n, cmp());
    cout<<pt[0].id<<" "<<pt[(n/2)].id<<endl;

    return 0;
}

Sort the intervals by length. and from i=0 to no_of_intervals, mark the query numbers.
Make sure that each query number is marked by only one interval.

Edited by author 17.05.2022 11:32

What's the test for WA#9?

Windows, linux, maxOS?
Why I can't include <windows.h>? (Compilation error: No such file or directory)

3 4
1   100 1000 0
1   1   1000 0
100 0   1000 100
//answer 1 1 2 2

1 1
1000000000
//answer 1

5 6
525   0 171   0 872 673
  0 843   0   0   0   0
  0 277   0 202   0   0
  0   0 733 957  65  96
637 566   0   0   0 441
//answer: 4 4 5 5 5 5

10 1  10 10 10
1  1  10 10 10
1  10 1  1  1
1  1  1  10 1
10 10 10 10 1
//answer: 2 2 1 1 1 1

1 10
100 90 80 70 60 50 40 30 20 10
//answer: 10

5 6
1 0 1 1 1 1
1 0 0 1 1 1
1 1 0 1 1 1
1 0 0 1 1 1
1 0 1 1 1 1
//answer: 2 2 3 3 3 2 2

3 2
3 1
1 1
1 3
//answer: 2 2 1 1

3 1
1
1
1
//answer: 1 1 1

тест 1: n=9 k=2
тест 2: n=1 k=1
тест 3: n=10 k=1
тест 4: n=8 k=1
тест 5: n=10 k=2
тест 6: n=7 k=2
тест 7: n=1 k=2
тест 8: n=2
тест 9: n=10 k=5
тест 10: n=9 k=5
n=3,4,5,6 не встречается...

clarification if you are stuck.
RedBlue Mana can be made from red or blue i.e C can use (remA + remB) after using A and B through X and Y respectively. Do count what you use from C when required.