/*
Given a linked list of the form, 1->2->3->4->5->6, convert it into the form 2->1->4->3->6->5. Note that the nodes need to be altered and not the data contained in them;
*/
#include<stdio.h>
#include<malloc.h>
struct node{
int a;
struct node *next;
}*start=NULL;
void add(int item)
{
struct node *tmp=(struct node *)malloc(sizeof(struct node));
struct node *tmp2;
tmp->a=item;
tmp->next=NULL;
tmp2=start;
if(tmp2!=NULL)
{
while(tmp2->next!=NULL) tmp2=tmp2->next;
tmp2->next=tmp;
}
else
start=tmp;
}
void display()
{
struct node *tmp=start;
while(tmp!=NULL){
printf("%d ",tmp->a);
tmp=tmp->next;
}
}
void reverse()
{
struct node *tmp=start;
struct node *tmp2=start->next;
struct node *tmp3=tmp2;
if(start->next==NULL) return;
tmp->next=NULL;
while(tmp2!=NULL)
{
tmp3=tmp2->next;
tmp2->next=tmp;
tmp=tmp2;
tmp2=tmp3;
}
start=tmp;
}
void swap()
{
struct node *tmp=start;
struct node *tmp2=tmp->next,*tmp3,*save;
if(tmp2==NULL) return;
if(tmp2->next==NULL)
{
tmp3=tmp;
tmp->next=tmp2->next;
tmp2->next=tmp;
start=tmp2;
return ;
}
if(tmp2->next->next==NULL)
{
tmp3=tmp2->next;
tmp->next=tmp3;
tmp2->next=tmp;
start=tmp2;
return ;
}
tmp3=tmp2->next;
tmp2->next=tmp;
tmp->next=tmp3->next;
start=tmp2;
tmp=tmp3;
tmp2=tmp->next;
while(1)
{
if(tmp2==NULL||tmp2->next==NULL||tmp2->next->next==NULL) break;
tmp3=tmp2->next;
if(tmp==NULL||tmp3->next==NULL) break;
tmp2->next=tmp;
tmp->next=tmp3->next;
tmp=tmp3;
tmp2=tmp->next;
}
if(tmp!=NULL&&tmp2!=NULL)
{
if(tmp2->next==NULL)
{
tmp2->next=tmp;
tmp->next=NULL;
}
else if(tmp2!=NULL)
{
save=tmp2->next;
tmp2->next=tmp;
tmp->next=save;
}
}
}
int main()
{
int item;
printf("enter the item one by one\nenter -1 to stop\n");
while(1)
{
item;
scanf("%d",&item);
if(item==-1) break;
add(item);
}
printf("\nThe single linked list is\n");
display();
reverse();
printf("\nThe reversed linked list is\n");
display();
printf("\nThe single linked list is\n");
reverse();
display();
printf("\nThe swapped(only 2 adjacent elements) linked list is\n");
printf("\nGiven a linked list of the form, 1->2->3->4->5->6, convert it into the form 2->1->4->3->6->5. Note that the nodes need to be altered and not the data contained in them.\n\n");
swap();
display();
return 0;
}
Interview question – Linked list June 1, 2011
Level Order Traversal(BFT) of a Binary Tree May 25, 2011
//level order traversal of a tree.
//here tree that is created is BST although it will work for every binary tree
#include<stdio.h>
#include<malloc.h>
#include<queue>
using namespace std;
struct node{
int item;
struct node *left;
struct node *right;
};
struct node *root=(struct node *)malloc(sizeof(struct node));
void insert(int data)
{
struct node *tmp=(struct node *)malloc(sizeof(struct node));
struct node *tmp2;
struct node *save=(struct node *)malloc(sizeof(struct node));
tmp->item=data;
tmp->right=NULL;
tmp->left=NULL;
tmp2=root;
while(1)
{
if(tmp2==NULL) break;
if(data>tmp2->item)
{
save=tmp2;
tmp2=tmp2->right;
}
else if(data<tmp2->item)
{
save=tmp2;
tmp2=tmp2->left;
}
}
if(save->item<data)
{
save->right=tmp;
}
else if(save->item>data)
{
save->left=tmp;
}
}
//////////////////level order traversal begins//////////////////////
void bfs()
{
queue<struct node *>q1;
queue<struct node *>q2;
struct node *tmp;
struct node *tmp2;
q1.push(root);//root
while(1)
{
while(!q1.empty())
{
tmp2=q1.front();
if(tmp2->left!=NULL)
{
q2.push(tmp2->left);
}
if(tmp2->right!=NULL)
{
q2.push(tmp2->right);
}
printf("%d ",tmp2->item);
q1.pop();
}
if(q2.empty()) break;
printf("\n");
while(!q2.empty())
{
q1.push(q2.front());
q2.pop();
}
}
}
//////////////////level order traversal end//////////////////////
int main()
{
int data;
//create binary search tree
printf("enter the root element");
scanf("%d",&data);
root->item=data;
root->left=NULL;
root->right=NULL;
while(1)
{
printf("enter -1 to stop");
scanf("%d",&data);
if(data==-1) break;
insert(data);
}
bfs();//level order traversal
return 0;
}
2010 in review January 2, 2011
The stats helper monkeys at WordPress.com mulled over how this blog did in 2010, and here’s a high level summary of its overall blog health:
The Blog-Health-o-Meter™ reads Fresher than ever.
Crunchy numbers
A helper monkey made this abstract painting, inspired by your stats.
A Boeing 747-400 passenger jet can hold 416 passengers. This blog was viewed about 8,900 times in 2010. That’s about 21 full 747s.
In 2010, there were 95 new posts, growing the total archive of this blog to 299 posts. There were 4 pictures uploaded, taking up a total of 157kb.
The busiest day of the year was May 4th with 134 views. The most popular post that day was SPOJ UBoat Problem code: HS09UBT.
Where did they come from?
The top referring sites in 2010 were en.wordpress.com, google.co.in, tausiq.wordpress.com, alliraph.wordpress.com, and google.com.
Some visitors came searching, mostly for c++ bash arena, “s1 = (char*)((int)s1 + (int)s2)”, how to unlock yum, anarc,spoj, and unlock yum.
Attractions in 2010
These are the posts and pages that got the most views in 2010.
1
SPOJ UBoat Problem code: HS09UBT February 2010
2
SPOJ 5842. Polybius square February 2010
1 comment
3
nfa to dfa conversion best explained July 2010
4
Compile Linux kernel fedora 11 January 2010
5
How to unlock yum in fedora February 2010
2 comments and 1 Like on WordPress.com,
Exception in thread “main” java.lang.NoClassDefFoundError: Main September 4, 2010
Sometimes , inspite of setting proper classpath for JAVA_HOME, getting this Error on running
$java filename(without exten) – Producing error means something strange. I too got irritated and finally found that its bcz its not considering the present working directory.
So by doing this
$java -classpath . filename
problem was solved
Here is the Great Hepful Link
Modular Exponentiation September 4, 2010
Problem5
CodeChef
ACCEPTED
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
using namespace std;
int main()
{
int A,B;
while(1)
{
scanf("%d%d",&A,&B);
if(A==0&&B==0) break;
int base,power,modulus;
modulus=1000000;
base=A;power=B;
long long result = 1;
for (int i = 31; i >= 0; i--) {
result = (result*result) % modulus;
if ((power & (1 << i)) != 0) {
result = (result*base) % modulus;
}
}
int ans=(int) result;
printf("%d\n",ans);
}
return 0;
}
Programming Quest June 29, 2010
From YoungProgrammer.com
LCM-GCD based question
Question:
We all interested in mathematics and world are divided in to two parts.
One who are interested in mathematics and other who are afraid of mathematics.
Here is a equation:
( X * N ) % Y = 0
Given two number X & Y you have to find minimum N that satisfies the equation.
Input:
Input consists of two positive integer X & Y . (1<=X,Y<=2000000000)
Output:
You have to output minimum N.
Sample Input:
1 5 6 7
Sample Output:
5
7
Note: You have to give the total of all output for each input.
For the sample input set, you have to give
7+5 = 12. So answer is 12.
Download data sets
ANS-562252
my solution
#include<cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <ctime>
#include <cstring>
#include <cassert>
#include <sstream>
#include <iomanip>
#include <complex>
#include <queue>
#include <functional>
using namespace std;
#define forn(i, n) for(int i = 0; i < (int)(n); i++)
#define ford(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define last(a) int(a.size() - 1)
#define all(a) a.begin(), a.end()
#define seta(a,x) memset (a, x, sizeof (a))
#define I (int)
#define SZ(x) ((int) (x).size())
#define FE(i,x) for(typedef((x).begin() i=(x).begin();i!=(x).end();i++)
typedef long long int int64;
typedef unsigned long long int uint64;
typedef long double ldb;
typedef pair <int, int> pii;
typedef vector<int>vi;
typedef vector<string>vs;
template <class T> T sqr (T x) {return x * x;}
int gcd(int a,int b)
{
if(b==0) return a;
return gcd(b,a%b);
}
int main()
{
int X,Y;
int total=0;
while(1)
{
cin>>X>>Y;
int hcf=gcd(X,Y);
int lcm=(X*Y)/hcf;
total+=(lcm/X);
cout<<total<<endl;
}
return 0;
}
Programming Quest June 29, 2010
If a number is the only number between a prime number and a square number it is
beprisque.
Such few numbers are: 2 3 8 10
2 is beprisque, because, 1 is a square and 3 is a prime number
3 is beprisque, because, 2 is prime and 4 is a square number
What's the 100th beprisque number ?
ANS-119026
From Young programmer.com
my solution (not optimized)
#include<cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <ctime>
#include <cstring>
#include <cassert>
#include <sstream>
#include <iomanip>
#include <complex>
#include <queue>
#include <functional>
using namespace std;
#define forn(i, n) for(int i = 0; i < (int)(n); i++)
#define ford(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define last(a) int(a.size() - 1)
#define all(a) a.begin(), a.end()
#define seta(a,x) memset (a, x, sizeof (a))
#define I (int)
#define SZ(x) ((int) (x).size())
#define FE(i,x) for(typedef((x).begin() i=(x).begin();i!=(x).end();i++)
typedef long long int int64;
typedef unsigned long long int uint64;
typedef long double ldb;
typedef pair <int, int> pii;
typedef vector<int>vi;
typedef vector<string>vs;
template <class T> T sqr (T x) {return x * x;}
int main()
{
static int prime[1000009];
static int sqr[1000009];
for(int i=1;i<=1000;i++)
sqr[i*i]=1;
prime[1]=0;prime[2]=1;
for(int i=3;i<=1000000;i=i+2)
{
int flag=1;
for(int j=3;j<=sqrt(i);j++)
{
if(i%j==0) { flag=0;break;}
else continue;
}
if(flag==1) prime[i]=1;
}
int count=0;
for(int i=2;i<=120000;i++)
{
if((sqr[i-1]==1&&prime[i+1]==1)||(sqr[i+1]==1&&prime[i-1]==1)){cout<<i<<endl; ++count;}
}
cout<<count<<endl;
return 0;
}
here is the few BEPRISQUE number
2
3
8
10
24
48
80
82
168
224
226
360
440
442
728
840
1088
1090
1224
1368
1522
1848
2026
2208
2400
3024
3250
3720
3968
4760
5040
5624
5928
6562
7920
8648
9802
10608
11026
11448
12322
13688
13690
14160
14640
15130
16128
17160
18224
19320
21024
21610
24024
25920
28560
29242
29928
31328
33488
36480
42024
44520
47088
47962
49728
50626
53360
54288
56168
56170
57120
59050
61008
62002
64008
65026
66048
67080
70224
71288
74528
74530
77840
81224
85848
88210
89400
90600
91808
91810
95480
95482
97968
99224
103042
104328
112224
113568
116280
119026
100
Programming Quest June 29, 2010
Determine the number of n-bit sequences that contain no adjacent 1’s. For
example, for n = 3 the answer is 5 (sequences 000, 001, 010, 100, 101 are
acceptable while 011, 110, 111 are not).
Find the answer where n = 10
ANS-144
#include<cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <ctime>
#include <cstring>
#include <cassert>
#include <sstream>
#include <iomanip>
#include <complex>
#include <queue>
#include <functional>
using namespace std;
#define forn(i, n) for(int i = 0; i < (int)(n); i++)
#define ford(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define last(a) int(a.size() - 1)
#define all(a) a.begin(), a.end()
#define seta(a,x) memset (a, x, sizeof (a))
#define I (int)
#define SZ(x) ((int) (x).size())
#define FE(i,x) for(typedef((x).begin() i=(x).begin();i!=(x).end();i++)
typedef long long int int64;
typedef unsigned long long int uint64;
typedef long double ldb;
typedef pair <int, int> pii;
typedef vector<int>vi;
typedef vector<string>vs;
template <class T> T sqr (T x) {return x * x;}
int main()
{
char s[1028];
int count=0;
for(int i=0;i<1024;i++)
{
int tmp=i;
int j=0;
if(i==0) {s[0]='0';s[1]='\0';}
while(tmp>0)
{
int rem=tmp%2;
s[j++]=rem+'0';
tmp=tmp/2;
}
if(i!=0)
s[j]='\0';
//printf("%s\n",s);
if(!strstr(s,"11")) ++count;
}
cout<<count<<endl;
return 0;
}
Programming Question June 29, 2010
A number has 6 different digits. If the last digit is moved to the front, then a new number is formed which is exaclty 5 times the old number. With which number did we start?
my solution
#include<cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <ctime>
#include <cstring>
#include <cassert>
#include <sstream>
#include <iomanip>
#include <complex>
#include <queue>
#include <functional>
using namespace std;
#define forn(i, n) for(int i = 0; i < (int)(n); i++)
#define ford(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define last(a) int(a.size() - 1)
#define all(a) a.begin(), a.end()
#define seta(a,x) memset (a, x, sizeof (a))
#define I (int)
#define SZ(x) ((int) (x).size())
#define FE(i,x) for(typedef((x).begin() i=(x).begin();i!=(x).end();i++)
typedef long long int int64;
typedef unsigned long long int uint64;
typedef long double ldb;
typedef pair <int, int> pii;
typedef vector<int>vi;
typedef vector<string>vs;
template <class T> T sqr (T x) {return x * x;}
int main()
{
char s[15];
int count[10];
for(int i=102345;i<=999999;i++)
{
sprintf(s,"%d",i);
for(int k=0;k<10;k++)
count[k]=0;
int flag=1;
for(int j=0;j<strlen(s);j++){
count[s[j]-'0']++;
if(count[s[j]-'0']>=2) {flag=0;break;}
else continue;
}
if(flag==1)
{
char str[15];
str[0]=s[strlen(s)-1];
for(int j=0;j<strlen(s)-1;j++)
str[j+1]=s[j];
str[strlen(s)]='\0';
int num=atoi(str);
if(num==5*i) printf("%d\n",i);
}
}
return 0;
}
//Answer is 142857
Fibonacci Or Not June 29, 2010
The condition to check whether the given number is fibonacci or not. Or say printing numbers which are fibonacci ..
Condition :
N is a Fibonacci number if and only if 5 N2 + 4 or 5 N2 – 4 is a square number.
This was given by I Gessel in 1972. Here is the link
program in C++
</div>
<div>//checking whether the number is fibonacci or not
#include<cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <ctime>
#include <cstring>
#include <cassert>
#include <sstream>
#include <iomanip>
#include <complex>
#include <queue>
#include <functional>
using namespace std;
#define forn(i, n) for(int i = 0; i < (int)(n); i++)
#define ford(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define last(a) int(a.size() - 1)
#define all(a) a.begin(), a.end()
#define seta(a,x) memset (a, x, sizeof (a))
#define I (int)
#define SZ(x) ((int) (x).size())
#define FE(i,x) for(typedef((x).begin() i=(x).begin();i!=(x).end();i++)
typedef long long int int64;
typedef unsigned long long int uint64;
typedef long double ldb;
typedef pair <int, int> pii;
typedef vector<int>vi;
typedef vector<string>vs;
template <class T> T sqr (T x) {return x * x;}
int main()
{
int count=0;
for(int64 i=0;i<1000000000/*2147483639*/;i++)
{
int64 N=i;
int64 tmp1=5*N*N+4,tmp2=5*N*N-4;//whether the number is fibonacci or not
ldb t1=(ldb)sqrt((ldb)tmp1);
ldb t2=(ldb)sqrt((ldb)tmp2);
int64 tt1=(int64)t1,tt2=(int64)t2;
if(t1==(ldb)tt1||t2==(ldb)tt2){
printf("%d %lld\n",++count,i);
}
}
return 0;
}</div>
<div></div>
<div>
Here is the few Fibonacci numbers
s.no fib
1 0
2 1
3 2
4 3
5 5
6 8
7 13
8 21
9 34
10 55
11 89
12 144
13 233
14 377
15 610
16 987
17 1597
18 2584
19 4181
20 6765
21 10946
22 17711
23 28657
24 46368
25 75025
26 121393
27 196418
28 317811
29 514229
30 832040
31 1346269
32 2178309
33 3524578
34 5702887
35 9227465
36 14930352
37 24157817
38 39088169
39 63245986
40 102334155
41 165580141
42 267914296
43 433494437
44 701408733
2 1
3 2
4 3
5 5
6 8
7 13
8 21
9 34
10 55
11 89
12 144
13 233
14 377
15 610
16 987
17 1597
18 2584
19 4181
20 6765
21 10946
22 17711
23 28657
24 46368
25 75025
26 121393
27 196418
28 317811
29 514229
30 832040
31 1346269
32 2178309
33 3524578
34 5702887
35 9227465
36 14930352
37 24157817
38 39088169
39 63245986
40 102334155
41 165580141
42 267914296
43 433494437
44 701408733
UVA Contest from BUBT, Bangladesh June 29, 2010
Solved 2 problems A and B during the contest.
#include<cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <ctime>
#include <cstring>
#include <cassert>
#include <iomanip>
#include <complex>
#include <queue>
#include <functional>
using namespace std;
#define forn(i, n) for(int i = 0; i < (int)(n); i++)
#define ford(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define last(a) int(a.size() - 1)
#define all(a) a.begin(), a.end()
#define seta(a,x) memset (a, x, sizeof (a))
#define I (int)
#define SZ(x) ((int) (x).size())
//#define FE(i,x) for(typedef(x).begin() i=(x).begin();i!=(x).end();i++)
typedef long long int int64;
typedef unsigned long long int uint64;
typedef long double ldb;
typedef pair <int, int> pii;
typedef vector<int>vi;
typedef vector<string> vs;
template <class T> T sqr (T x) {return x * x;}
int main()
{
int cases;
cin>>cases;
char s[100];
vector<string>vs;
int a[11],b[11];
int c=0;
while(cases--)
{
vs.clear();
for(int i=0;i<10;i++)
{
scanf("%s%d%d",s,&a[i],&b[i]);
string str=(string)(s);
vs.pb(str);
}
for(int i=0;i<10;i++)
{
for(int j=0;j<10;j++)
{
if(a[i]<a[j])
{
swap(a[i],a[j]);swap(vs[i],vs[j]);swap(b[i],b[j]);
}
else if(a[i]==a[j])
{
if(b[i]>b[j])
{
swap(b[i],b[j]);swap(vs[i],vs[j]);swap(a[i],a[j]);
}
else if(b[i]==b[j])
{
if(vs[i]>vs[j])
{
swap(b[i],b[j]);swap(vs[i],vs[j]);swap(a[i],a[j]);
}
}//else
}//else
}//for
}//for
sort(vs.begin(),vs.begin()+5);
sort(vs.begin()+5,vs.end());
printf("Case %d:\n",++c);
printf("(");
for(int i=5;i<9;i++)
cout<<vs[i]<<", ";
cout<<vs[9]<<")\n";
printf("(");
for(int i=0;i<4;i++)
cout<<vs[i]<<", ";
cout<<vs[4]<<")\n";
}
return 0;
}</pre>
<pre>
Problem B Bafana Bafana
#include<cstdlib>
#include <cstdio>
#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <ctime>
#include <cstring>
#include <cassert>
#include <iomanip>
#include <complex>
#include <queue>
#include <functional>
using namespace std;
#define forn(i, n) for(int i = 0; i < (int)(n); i++)
#define ford(i, n) for(int i = (int)(n) - 1; i >= 0; i--)
#define pb push_back
#define mp make_pair
#define fs first
#define sc second
#define last(a) int(a.size() - 1)
#define all(a) a.begin(), a.end()
#define seta(a,x) memset (a, x, sizeof (a))
#define I (int)
#define SZ(x) ((int) (x).size())
#define FE(i,x) for(typedef(x).begin() i=(x).begin();i!=(x).end();i++)
typedef long long int int64;
typedef unsigned long long int uint64;
typedef long double ldb;
typedef pair <int, int> pii;
typedef vector<int>vi;
typedef vector<string>vs;
template <class T> T sqr (T x) {return x * x;}
int main()
{
int cases;
cin>>cases;
int N,K,P;
int i=0;
while(cases--)
{
cin>>N>>K>>P;
int tmp=(K+P)%N;
printf("Case %d: ",++i);
if(tmp==0) cout<<N<<endl;
else cout<<tmp<<endl;
}
return 0;
}
UVA Contest from Dinajpur, Bangladesh June 29, 2010
LINK
I solved a single problem during this UVA contest
F Horror Dash
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstdio>
using namespace std;
int main()
{
int cases;
cin>>cases;
int i=0;
while(cases--)
{
int n;
cin>>n;
cout<<"Case "<<++i<<": ";
int max=-1;
while(n--){
int item;
cin>>item;
if(item>max) max=item;
}
cout<<max;
if(cases!=0) printf("\n");
}
return 0;
}
TJU 3181. Stock Exchange June 10, 2010
TJU 3181. Stock Exchange
ACCEPTED
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<cmath>
using namespace std;
int main()
{
char s[100],s1[100],s2[100];
int i,j,k;
vector<double>price;
vector<string>agent,buysell;
while(1){
int n;
scanf("%d%s",&n,s);
string str=s,str1;
double d;
if(n==0&&str=="END") break;
agent.clear();buysell.clear();price.clear();
for(int i=0;i<n;i++){
scanf("%s%s%lf",s1,s2,&d);
str1=s1;agent.push_back(str1);
str1=s2;buysell.push_back(str1);
price.push_back(d);
}
vector< vector<string> > v;
vector<string> tmp;
vector< vector<string> >::iterator x;
//for(x=v.begin();x<v.begin();x++)
// (*x).clear();
v.clear();
tmp.clear();
vector<string>::iterator it1,it2,it3,it4;
it3=buysell.begin();it4=buysell.begin();
vector<double>::iterator it5,it6;
it5=price.begin();it6=price.begin();
for(it1=agent.begin();it1<agent.end();it1++,it3++,it5++){
//cout<<"yes";
tmp.clear();
int flag=0;
for(it2=agent.begin(),it4=buysell.begin(),it6=price.begin();it2<agent.end();it2++,it4++,it6++){
if(it1!=it2){
if(*it3!=*it4){
if(*it3=="buy"){
if(*it6<=*it5){
tmp.push_back(*it2);
flag=1;
}
}
else if(*it3=="sell"){
if(*it5<=*it6){
tmp.push_back(*it2);
flag=1;
}
}
}
}
}
if(flag==0) {tmp.push_back("NO-ONE");
v.push_back(tmp);
}
else
v.push_back(tmp);
}
printf("%s\n",s);
vector<string>::iterator it=agent.begin();
vector< vector<string> >::iterator ii;
vector<string>::iterator ii2;
ii=v.begin();
for(int i=0;i<n;i++,it++,ii++){
cout<<*it<<":";
for(ii2=(*ii).begin();ii2<(*ii).end();ii2++){
cout<<" "<<(*ii2);
}
cout<<endl;
}
}
return 0;
}
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