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Обсуждение задачи 1149. Танцы синуса

Easiest way to solve this problem [C++ code with command].
Послано Akash 25 авг 2020 00:35
/***************************************************************************************************
                         Anzmaul Haque Akash
                    DIU, Daffodil International University
                    E-mail: anzamul15-2297@diu.edu.bd
 Youtube:https://www.youtube.com/channel/UCXvXHUkxyLYKeQkIg_RyMBw
******************************************************************************************************/
#include<bits/stdc++.h>

#define ln     (int)1e6
#define pi     acos(-1.0)
#define pb     push_back
#define pf     push_front 1
#define mp     make_pair
#define F      first
#define S      second
#define INF    1000000000000000000
#define LES    -1000000000000000000
#define intlim 2147483648
#define ll     long long
#define ull    unsigned long long
#define str    string
#define in     insert

#define all(akash)       akash.begin(),akash.end()
#define ms(arr)          memset(arr,0,sizeof(arr))
#define msbooltrue(arr)  memset(arr,true,sizeof(arr))
#define msboolfalse(arr) memset(arr,false,sizeof(arr))
#define msINF(arr)       memset(arr,INF,sizeof(arr))
#define case1(cs)        cout<<"Case "<<cs<<": "
#define case2(cs)        cout<<"Case "<<cs<<":\n"
#define lb               cout<<"\n";

#define vec vector
#define qu  queue
#define li  list
#define pq  priority_queue
#define em  empty

#define fori(i,a,b) for(int i=0;i<b;i++)
#define ford(i,a,b) for(int i=b-1;i>=0;i--)
#define forc(ch);   for(char ch:str)

#define sc1(a)         scanf("%lld",&a)
#define sc2(a,b)       scanf("%lld %lld",&a,&b)
#define sc3(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define sc4(a,b,c,d)   scanf("%lld %lld %lld %lld",&a,&b,&c,&d)
#define sc5(a,b,c,d,e) scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&e)

using namespace std;

void solve(ll n){

    ll temp_n=n;

    temp_n--;

    for(ll i=0; i<n-1; i++){
        cout<<"(";
    }
    cout<<"sin(1)+"<<n<<")"; //Segment 1.

    str s="sin(1-sin("; // if n>1 this one is using for the template.


    for(ll i=2;i<=n;i++){ //main loop.

        str temp;
        temp.clear();
        temp=s;


        for(ll x=2; x<i;x++){ //This one creat middle of Every part.

            if(x%2==0){ //' + ' for even ' - ' for odd.
                temp+=to_string(x);
                temp+="+sin(";
            }
            else{
                temp+=to_string(x);
                temp+="-sin(";
            }

        }

        temp+=to_string(i);

        for(ll y=1;y<=i;y++){ //for closing ' ) ' every equation.
            temp+=")";
        }

        cout<<temp; //print each segment.

        if(temp_n > 1){
            cout<<"+"<<temp_n<<")"; //Printing every segment last digit with out 1.
        }
        temp_n--;
    }

    cout<<"+1";lb//This one for end.
}

int main()
{
    ll n;
    sc1(n);

    if(n>1){
        solve(n); //solve function.
    }
    else if(n==1){
        cout<<"sin(1)+1";lb //for 1.
    }

    return 0;
}



















Re: Easiest way to solve this problem [C++ code with command].
Послано Akash 25 авг 2020 00:41
0
1
sin(1)+1
2
(sin(1)+2)sin(1-sin(2))+1
3
((sin(1)+3)sin(1-sin(2))+2)sin(1-sin(2+sin(3)))+1
4
(((sin(1)+4)sin(1-sin(2))+3)sin(1-sin(2+sin(3)))+2)sin(1-sin(2+sin(3-sin(4))))+1
5
((((sin(1)+5)sin(1-sin(2))+4)sin(1-sin(2+sin(3)))+3)sin(1-sin(2+sin(3-sin(4))))+2)sin(1-sin(2+sin(3-sin(4+sin(5)))))+1
6
(((((sin(1)+6)sin(1-sin(2))+5)sin(1-sin(2+sin(3)))+4)sin(1-sin(2+sin(3-sin(4))))+3)sin(1-sin(2+sin(3-sin(4+sin(5)))))+2)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6))))))+1
7
((((((sin(1)+7)sin(1-sin(2))+6)sin(1-sin(2+sin(3)))+5)sin(1-sin(2+sin(3-sin(4))))+4)sin(1-sin(2+sin(3-sin(4+sin(5)))))+3)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6))))))+2)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7)))))))+1
8
(((((((sin(1)+8)sin(1-sin(2))+7)sin(1-sin(2+sin(3)))+6)sin(1-sin(2+sin(3-sin(4))))+5)sin(1-sin(2+sin(3-sin(4+sin(5)))))+4)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6))))))+3)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7)))))))+2)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8))))))))+1
9
((((((((sin(1)+9)sin(1-sin(2))+8)sin(1-sin(2+sin(3)))+7)sin(1-sin(2+sin(3-sin(4))))+6)sin(1-sin(2+sin(3-sin(4+sin(5)))))+5)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6))))))+4)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7)))))))+3)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8))))))))+2)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8+sin(9)))))))))+1
10
(((((((((sin(1)+10)sin(1-sin(2))+9)sin(1-sin(2+sin(3)))+8)sin(1-sin(2+sin(3-sin(4))))+7)sin(1-sin(2+sin(3-sin(4+sin(5)))))+6)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6))))))+5)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7)))))))+4)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8))))))))+3)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8+sin(9)))))))))+2)sin(1-sin(2+sin(3-sin(4+sin(5-sin(6+sin(7-sin(8+sin(9-sin(10))))))))))+1
Re: Easiest way to solve this problem [C++ code with command].
Послано codeJutsu 1 ноя 2025 20:50
wtf is this code lol

also, why share code in the discussion section you shame of bangladesh!!! you're supposed to discuss here, not share code! also this problem is very easy, was no need to share code.