SPOJ 5676. STONE GAME
Problem code: RESN04
–AC–
#include
int main()
{
int T,n,i,j,count;
scanf(“%d”,&T);
while(T–){
scanf(“%d”,&n);
count=0;
for(i=1;i<=n;i++)
{
scanf("%d",&j);
if(i==j) ++count;
}
if(count%2==0) printf("BOB\n");
else printf("ALICE\n");
}
return 0;
}
TJU 2498. Uncle Jack
TODOLIST
WRONG ANSWER
{big multiplication based question}
#include<iostream>
#include<stdlib.h>
#include<math.h>
using namespace std;
class A{
long long i,j,N,D,R;
public:
void input(){
cin>>N>>D;
if(N==0&&D==0) exit(0);
}
void output(){
R=powf(N,D);
cout<<R<<endl;
}
};
int main()
{
A a;
while(1){
a.input();
a.output();
}
return 0;
}
#include<iostream>
using namespace std;
class A{
int i,j,l,n,m,a[201],b[201];
public:
void input(){
cin>>n;
for(i=0;i<n;i++)
cin>>a[i];
cin>>m;
for(i=0;i<m;i++)
cin>>b[i];
l=0;
for(i=0;i<n;i++)
for(j=0;j<m;j++)
if(a[i]+b[j]==0)
++l;
}
void output(){
cout<<l<<endl;
}
};
int main()
{
int cases;
cin>>cases;
A obj;
while(cases--)
{
obj.input();
obj.output();
}
return 0;
}
#include<iostream>
using namespace std;
class A{
int a[11],b[11],n,i,j;
public:
void input(){
cin>>n;
j=0;
for(i=0;i<n;i++)
cin>>a[i];
for(i=0;i<n;i++){
cin>>b[i];
j=j+a[i]*b[i];
}
}
void output(){
cout<<j<<endl;
}
};
int main()
{
A obj;
int cases;
cin>>cases;
while(cases--)
{
obj.input();
obj.output();
}
return 0;
}
SPOJ 5872. Anagram
Problem code: ANAG
c++/oops implemented
#include<iostream>
#include<string.h>
#include<stdlib.h>
using namespace std;
class A{
char s1[100],s2[100];
int i,j,k;
public:
void input(){
cin>>s1>>s2;
}
void cal(){
int a[26],b[26];
for(i=0;i<26;i++)
a[i]=b[i]=0;
for(i=0;s1[i]!='\0';i++)
{
a[s1[i]-'a']++;
}
for(i=0;s2[i]!='\0';i++)
{
b[s2[i]-'a']++;
}
output(a,b);
}
void output(int a[],int b[]){
int flag=0;
for(i=0;i<26;i++)
{
if(a[i]==b[i]){
flag=1;
continue;
}
else{
flag=0;
break;
}
}
if(flag==1) cout<<"YES\n";
else cout<<"NO\n";
}
};
int main()
{
A obj;
int cases;
cin>>cases;
while(cases--)
{
obj.input();
obj.cal();
}
return 0;
}
SPOJ 5848. Filip
Problem code: FILIP
C++ implemented
#include<iostream>
#include<stdlib.h>
using namespace std;
class A{
char s1[10],s2[10],c;
int i,j,k;
public:
void input(){
cin>>s1>>s2;
}
void cal(){
c=s1[0];
s1[0]=s1[2];
s1[2]=c;
c=s2[0];
s2[0]=s2[2];
s2[2]=c;
}
void output(){
int n1=atoi(s1);
int n2=atoi(s2);
if(n1>n2)
cout<<n1<<endl;
else
cout<<n2<<endl;
}
};
int main()
{
A a;
a.input();
a.cal();
a.output();
return 0;
}
Cryptography practise
TJU 3016. Decode the Strings
TODO[medium]
--AC--
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
int i,j,k;
int cases,cs=0;
char s[1000];
scanf("%d",&cases);
while(cases--)
{
++cs;
scanf("%d %s",&i,s);
printf("%d ",cs);
for(j=0;s[j]!='\0';j++)
if(i-1!=j)
printf("%c",s[j]);
printf("\n");
}
return 0;
}
#include<stdio.h>
#include<stdlib.h>
int main()
{
int i,j,k;
int n,v,M,J;
while(1)
{
scanf("%d",&n);
if(n==0) break;
M=0;J=0;
for(i=0;i<n;i++)
{
scanf("%d",&v);
if(v==0)
++M;
else
++J;
}
printf("Mary won %d times and John won %d times\n",M,J);
}
return 0;
}
SPOJ 1028. Hubulullu
Problem code: HUBULLU
Hard and Easy, depends which way u think
–AC–(easy way)
SPOJ 2528. Monkey Vines
Problem code: GNY07F
--AC--
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
int main()
{
int i,j,k;
int N,max,test=0;
char s[200],c;
scanf("%d",&N);
getchar();
while(N--)
{
i=0;
while(1)
{
c=getchar();
if(c=='\n')
{
s[i]='\0';
break;
}
s[i++]=c;
}
++test;
j=0;
if(s[0]=='\0'||s[0]=='\n')
max=1;
else
max=0;
for(i=0;s[i]!='\0';i++)
{
if(s[i]=='[')
j++;
else if(s[i]==']')
j--;
if(j>max)
max=j;
}
if(max==1&&s[0]=='\0')
printf("%d %d\n",test,max);
else
printf("%d %d\n",test,(int)pow(2,max));
}
return 0;
}
--AC--
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
int main()
{
int i,j,k;
int n,u,d,h,tmp,result,rem;
while(1)
{
scanf("%d%d%d",&n,&u,&d);
if(n==0) break;
i=u-d;
tmp=(n-u)/i;
rem=(n-u)%i;
if(rem!=0)
tmp=tmp+1;
result=2*tmp+1;
printf("%d\n",result);
}
return 0;
}
SPOJ 1025. Fashion Shows
Problem code: FASHION
–AC–
sorting based problem(qsort implemnted)
TJU 3288. Stockholm Numbers
original thinking implemented
1->1*2+1=3 //1 has odd parity
2->2*2+1=5//2has odd parity
3->3*2=6//3has even parity
4->4*2+1=9//4has odd parity
5->5*2=10//5has even parity
and so on
……………….
--AC--
#include<stdio.h>
int main()
{
long long int i,j,k,K,tmp,rem,cases;
scanf("%lld",&cases);
while(cases--)
{
scanf("%lld",&K);
j=0;
tmp=K;
while(K>0)
{
rem=K%2;
if(rem==1)
++j;
K=K/2;
}
if(j%2==0)
printf("%lld\n",2*tmp);
else
printf("%lld\n",2*tmp+1);
}
return 0;
}
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