Algelementari.docx
This document was uploaded by user and they confirmed that they have the permission to share it. If you are author or own the copyright of this book, please report to us by using this DMCA report form. Report DMCA
Overview
Download & View Algelementari.docx as PDF for free.
More details
- Words: 621
- Pages: 4
#include #include using namespace std; //nr prim int prim (int x) { if(x<2) return 0; if(x%2==0&&x!=2) return 0; for(int d=3;d*d<=x;d=d+2) { if(x%d==0) return 0; } return 1; } int cmmdc_s(int a,int b) { while(a!=b) { if(a>b) a=a-b; else b=b-a; } return a; } int cmmdc_euclid(int a,int b) { int r; while(b!=0) { r=a%b; a=b; b=r; } return a; }
// 121, 11 int palindrom(int x) { int aux,og=0; aux=x; while(aux>0) { og=og*10+(aux%10); aux=aux/10; } if(x==og) return 1; return 0; } // 25,36,81 int patratperfect(int x) { int r; r=sqrt(x); if(r*r==x) return 1; return 0; } //x=29 2+9=11 1+1=2; int cifra_control(int x) { if(x%9==0) return 9; //9,18,27 return x%9; } //n=100;s=1+2+4+5+10+20+25+50+100=217 long long sumdiv(int x) { long long s=0,d; for(d=1;d*d<x;d++) { if(x%d==0) s=s+d+x/d; } if(d*d==x) s=s+d; return s; }
//descompunere in factori primi //suma divizorilor primi //n=100 s=2+5=7 int sumdiv_prim(int x) { int s=0,d=2,exp; while(x>1) { exp=0; while(x%d==0) { exp++; x=x/d; } if(exp!=0) s=s+d; if(d==2) d=3; else d=d+2; } return s; } //n=1232 c=2 (dublam cifra c) m=122322 //vom construi un alt nr m int dublare(int x,int c) { int m=0,p=1; while(x>0) { m=(x%10)*p+m; p=p*10; if(c==x%10) { m=(x%10)*p+m; p=p*10; } x=x/10; } return m; }
//n=1232 c=2 (dupa eliminare) m=13 int elim_cif(int x,int c) { int m=0,p=1; while(x>0) { if(c!=x%10) { m=(x%10)*p+m; p=p*10; } x=x/10; } return m; } //numarul de 0-uri de la sfarsitul lui n! //25! are 6 0-uri la sfarsit n/5+n/25+n/125…. int nr_zero(int x) { int p=5,s=0; while(p<=x) { s=s+x/p; p=p*5; } return s; } //al n-lea termen al sirului Fibonacci //n=8 1 1 2 3 5 8 13 21 int fibo (int n) { int a=1,b=1,c; if(n<2) return 1; for(int i=3;i<=n;i++) { c=a+b; a=b; b=c; } return c; } //nr de 1 in baza 2 //4 este 100 in baza 2 int baza2 (int n) { int nr=0;
while(n>0) { if(n%2==1) nr++; n=n/2; } return nr; } //citire vector //n=5 28 5 -7 4 32 void citire (int &n, int v[]) { int i; cin>>n; for(i=1;i<=n;i++) cin>>v[i]; } //citim pana la aparitia lui 0 void citire(int &n, int v[]) { n=0; int x; cin>>x; while(x!=0) { v[++n]=x; cin>>x; } } void afisare (int n,int v[]) { int i; for(i=1;i<=n;i++) cout<
for(j=i+1;j<=n;j++) { if(v[i]>v[j]) { aux=v[i]; v[i]=v[j]; v[j]=aux; } } } void sort_buble (int n,int v[]) { int ok,i; do{ ok=1; for(i=1;ix)
{
{ v[j+1]=v[j]; j--;
int i=1,j=1; n=0; while(i<=n1&&j<=n2) { if(v1[i]
} v[j+1]=x; n++; } //separarea elementelor pare de cele impare void par_impar (int n,int v[]) { int i=1,j=n,aux; while(i<=j) { while(v[i]%2==0&&i<=n) i++; while(v[j]%2!=0&&j>=0) j--; if(i<=j) { aux=v[i]; v[i]=v[j]; v[j]=aux; i++; j--; } } } void secventa_para (int n,int v[],int &inc, int &sf, int &lmax) { int lc=0,lmax,i; for(i=1;i<=n;i++) { if(v[i]%2==0) lc++; else lc=0; if(lc>lmax) { lmax=lc; sf=i; inc=i-lc+1; } } } //cei doi vectori trebuie sa fie sortati crescator void interclasare (int n1,int v1[],int n2,int v2[],int &n,int v[])
} //primul vector este sortat crescator, iar al doilea descrescator void interclasare (int n1,int v1[],int n2,int v2[],int &n,int v[]) { int i=1,j=n2; n=0; while(i<=n1&&j>=1) { if(v1[i]=1) v[++n]=v2[j--]; } int main() { return 0; }
// 121, 11 int palindrom(int x) { int aux,og=0; aux=x; while(aux>0) { og=og*10+(aux%10); aux=aux/10; } if(x==og) return 1; return 0; } // 25,36,81 int patratperfect(int x) { int r; r=sqrt(x); if(r*r==x) return 1; return 0; } //x=29 2+9=11 1+1=2; int cifra_control(int x) { if(x%9==0) return 9; //9,18,27 return x%9; } //n=100;s=1+2+4+5+10+20+25+50+100=217 long long sumdiv(int x) { long long s=0,d; for(d=1;d*d<x;d++) { if(x%d==0) s=s+d+x/d; } if(d*d==x) s=s+d; return s; }
//descompunere in factori primi //suma divizorilor primi //n=100 s=2+5=7 int sumdiv_prim(int x) { int s=0,d=2,exp; while(x>1) { exp=0; while(x%d==0) { exp++; x=x/d; } if(exp!=0) s=s+d; if(d==2) d=3; else d=d+2; } return s; } //n=1232 c=2 (dublam cifra c) m=122322 //vom construi un alt nr m int dublare(int x,int c) { int m=0,p=1; while(x>0) { m=(x%10)*p+m; p=p*10; if(c==x%10) { m=(x%10)*p+m; p=p*10; } x=x/10; } return m; }
//n=1232 c=2 (dupa eliminare) m=13 int elim_cif(int x,int c) { int m=0,p=1; while(x>0) { if(c!=x%10) { m=(x%10)*p+m; p=p*10; } x=x/10; } return m; } //numarul de 0-uri de la sfarsitul lui n! //25! are 6 0-uri la sfarsit n/5+n/25+n/125…. int nr_zero(int x) { int p=5,s=0; while(p<=x) { s=s+x/p; p=p*5; } return s; } //al n-lea termen al sirului Fibonacci //n=8 1 1 2 3 5 8 13 21 int fibo (int n) { int a=1,b=1,c; if(n<2) return 1; for(int i=3;i<=n;i++) { c=a+b; a=b; b=c; } return c; } //nr de 1 in baza 2 //4 este 100 in baza 2 int baza2 (int n) { int nr=0;
while(n>0) { if(n%2==1) nr++; n=n/2; } return nr; } //citire vector //n=5 28 5 -7 4 32 void citire (int &n, int v[]) { int i; cin>>n; for(i=1;i<=n;i++) cin>>v[i]; } //citim pana la aparitia lui 0 void citire(int &n, int v[]) { n=0; int x; cin>>x; while(x!=0) { v[++n]=x; cin>>x; } } void afisare (int n,int v[]) { int i; for(i=1;i<=n;i++) cout<
for(j=i+1;j<=n;j++) { if(v[i]>v[j]) { aux=v[i]; v[i]=v[j]; v[j]=aux; } } } void sort_buble (int n,int v[]) { int ok,i; do{ ok=1; for(i=1;i
{
{ v[j+1]=v[j]; j--;
int i=1,j=1; n=0; while(i<=n1&&j<=n2) { if(v1[i]
} v[j+1]=x; n++; } //separarea elementelor pare de cele impare void par_impar (int n,int v[]) { int i=1,j=n,aux; while(i<=j) { while(v[i]%2==0&&i<=n) i++; while(v[j]%2!=0&&j>=0) j--; if(i<=j) { aux=v[i]; v[i]=v[j]; v[j]=aux; i++; j--; } } } void secventa_para (int n,int v[],int &inc, int &sf, int &lmax) { int lc=0,lmax,i; for(i=1;i<=n;i++) { if(v[i]%2==0) lc++; else lc=0; if(lc>lmax) { lmax=lc; sf=i; inc=i-lc+1; } } } //cei doi vectori trebuie sa fie sortati crescator void interclasare (int n1,int v1[],int n2,int v2[],int &n,int v[])
} //primul vector este sortat crescator, iar al doilea descrescator void interclasare (int n1,int v1[],int n2,int v2[],int &n,int v[]) { int i=1,j=n2; n=0; while(i<=n1&&j>=1) { if(v1[i]
More Documents from "SerenaBlaga"
Cautarebinara_aplicatii.pdf
November 2019 3
Algelementari.docx
June 2020 2
Info-1.docx
June 2020 5
Cat De Postmodernist Sunt.docx
November 2019 3
Eficienta.docx
June 2020 5