by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <stdio.h>
int main()
{
int n, i, q, c, sz, idx, idx2, t1, t2, elteres, ce = 0;
while(1)
{
scanf("%d",&n);
if(!n) break;
printf("Case %d:\n",++ce);
int set[n], i = 0, q = 0;
sz = n;
while(n--)
{
scanf("%d",&set[i++]);
}
scanf("%d",&q);
while(q--)
{
scanf("%d",&c);
elteres = 2147483647;
for(idx = 0; idx < sz; idx++)
{
for(idx2 = idx + 1; idx2 < sz; idx2++)
{
if( abs((set[idx] + set[idx2]) - c) < elteres )
{
elteres = abs((set[idx] + set[idx2]) - c);
t1 = set[idx];
t2 = set[idx2];
}
}
}
printf("Closest sum to %d is %d.\n",c,t1+t2);
}
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int n;
int prime[] = {2,3,5,7,11,13,17,19,23,29,31};
int sum,i,j,count;
while(scanf("%d",&n)==1)
{
if(n==0)
break;
count =0;
for(i=0; i<11; i++)
{
if(n==prime[i])
count =1;
}
if(count==1)
{
if(n==11 || n==23 || n==29)
{
cout<<"Given number is prime. But, NO perfect number is available." << endl;
}
else
{
sum = pow(2,n-1) * (pow (2,n)-1);
cout << "Perfect: " << sum <<'!'<<endl;
}
}
else
{
cout << "Given number is NOT prime! NO perfect number is available."<<endl;
}
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <bits/stdc++.h>
using namespace std;
int main()
{
long int r ,a;
while(scanf("%ld",&r)==1)
{
if(r<0) break;
if(r==0) printf("0%\n");
else if(r==1) printf("0%\n");
else if(r>=2)
{
a=r*25;
printf("%ld%%\n",a);
}
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include<bits/stdc++.h>
using namespace std;
int main()
{
int i,j,k,d,sum,num,tc,n,ans,digit;
while(cin>>tc)
{
for(k=1;k<=tc;k++)
{
cin>>n;
num=n,ans=0;
while(n>9)
{
while(n!=0)
{
digit=n%10;
n/=10;
ans+=digit*digit;
}
n=ans;
ans=0;
}
while(n<9)
{
if(n<=9){
ans=n;
break;
}
}
if(ans==1 || ans==7)
printf("Case #%lld: %lld is a Happy number.\n",k,num);
else
printf("Case #%lld: %lld is an Unhappy number.\n",k,num);
}
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include<bits/stdc++.h>
using namespace std;
int main()
{
unsigned long long n,N;
while(cin>>n and n>0)
{
N=(n*10)/9;
if(n%9==0)
{
cout<<N-1<<" "<<N<<endl;
}
else
{
cout<<N<<endl;
}
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <bits/stdc++.h>
using namespace std;
int a[10000];
int main()
{
int n,t,i,j;
while(scanf("%d",&t)==1)
{
j=0;
if(t==0)
return 0;
for(i=0;i<t;i++)
{
scanf("%d",&a[i]);
if(a[i] > 0)
{
j++;
a[n++]=a[i];
}
}
for(i=0;i<n;i++)
{
if(j==1)
printf(" ");
j=1;
printf("%d",a[i]);
}
if (j == 0)
printf("0");
printf("\n");
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <stdio.h>
int main()
{
int n;
while((scanf("%d",&n)==1) && n!=0)
{
if(n>=101)
printf("f91(%d) = %d\n",n,n-10);
else
printf("f91(%d) = 91\n",n);
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include<bits/stdc++.h>
bool flag[1000005];
int List[1000005],primes[100000];
int listSize,cnt;
using namespace std;
void sieve(int n)
{
cnt=0;
primes[cnt++] = 2;
for(int i=3; i<=n; i+=2)
{
if(flag[i] == 0)
{
primes[cnt++] = i;
if(i <= n/i)
{
for(int j=i*i; j<=n; j+=i*2)
flag[j] = 1;
}
}
}
return ;
}
void primeFactorize( int n )
{
listSize = 0;
for( int i = 0;i<cnt and primes[i]<=(int)sqrt(n);i++ )
{
if( n % primes[i] == 0 )
{
listSize++;
while( n % primes[i] == 0 )
{
n /= primes[i];
//List[listSize] = primes[i];
}
}
}
//if( n ==1 )
// break;
//List[listSize] = n;
//listSize++;
return ;
}
int main()
{
int n,N;
sieve(1000005);
while(cin>>n)
{
if(n==0) break;
if(n==1)
{
cout<<"0"<<endl;
continue;
}
N=n;
primeFactorize(N);
cout<<N<<" : "<<listSize+1<<endl;
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <stdio.h>
#include <string.h>
int main()
{
char s1[500],s2[500],s1s[500],s2s[500];
int t,i,j,k;
scanf("%d ",&t);
for(i=0;i<t;i++){
gets(s1);
gets(s2);
if(strcmp(s1,s2)==0)printf("Case %d: Yes\n",i+1);
else {
k=0;
for(j=0;j<strlen(s1);j++){
if(s1[j]!=' '){
s1s[k]=s1[j];
k++;
}
}
s1s[k]='\0';
k=0;
for(j=0;j<strlen(s2);j++){
if(s2s[j]!=' '){
s2s[k]=s2[j];
k++;
}
}
s2s[k]='\0';
if(strcmp(s1s,s2s)==0)printf("Case %d: Output Format Error\n",i+1);
else printf("Case %d: Wrong Answer\n",i+1);
}
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include<bits/stdc++.h>
using namespace std;
int main(){
double d,v,u,t1,t2;
int t;
cin>>t;
for(int i=0;i<t;i++){
cin>>d>>v>>u;
if( u==0 || v==0 || v>=u ){
printf("Case %d: can't determine\n",i+1);
continue;
}
t1=d/u;
t2=d/sqrt(u*u-v*v);
printf("Case %d: %.3lf\n",i+1,(t2-t1));
}
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <stdio.h>
int main()
{
int t,a,b,i,j=1,sum;
scanf("%d",&t);
while( t-- )
{
sum = 0;
scanf("%d %d",&a,&b);
for( i=a ; i<=b ; i++)
{
if(i%2==1)
sum += i;
}
printf("Case %d: %d\n",j,sum);
j++;
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <bits/stdc++.h>
using namespace std;
int main()
{
double N,n;
int c=1;
//long int n;
while(cin>>N and N!=0)
{
n=ceil((sqrt(8*N + 9)+ 3)/2);
printf("Case %d: %.lf\n",c++,n);
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include<bits/stdc++.h>
using namespace std;
int main()
{
long long int a,b,i=1;
while(cin>>a>>b)
{
if(a==0 && b==0) break;
if(a==0 || b==0)
{
printf("Case %lld: 0\n",i);
continue ;
}
printf("Case %lld: %lld\n",i++,((a*b*(a-1)*(b-1))/4));
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <stdio.h>
#include<iostream>
#include<math.h>
#define MAX 10001
using namespace std;
bool flag[MAX];
int primes[MAX];
int cnt;
void sieve(int n)
{
cnt=0;
primes[cnt++] = 2;
for(int i=3; i<=n; i+=2)
{
if(flag[i] == 0)
{
primes[cnt++] = i;
if(i <= n/i)
{
for(int j=i*i; j<=n; j+=i*2)
flag[j] = 1;
}
}
}
return ;
}
int main()
{
int tc, n;
sieve(MAX);
scanf("%d",&tc);
while ( tc-- )
{
scanf("%d",&n);
int ans = primes[0];
for (int i=0; i<cnt && primes[i]<=n; i++)
if ( n % primes[i] > n % ans )
printf("v=%d\tm=%d\n",n % primes[i],n % ans);
ans = primes[i];
/// printf("%d\n",ans);
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <iostream>
#include <algorithm>
#include <set>
#include <cmath>
using namespace std;
int main(void) {
int tc, c, r,t,q;
set<int> v;
set<int>::iterator it;
cin >> tc;
for (t = 0; t < tc; t++) {
cin >> c >> r;
v.clear();
cout << "Case #" << t+1 << ":";
if (c == r) {
cout << " 0" << endl;
continue;
}
q = c - r;
for (int i = 1; i <= sqrt(q); i++) {
if (q % i == 0) {
if (i > r)
v.insert(i);
if (q / i > r)
v.insert(q/i);
}
}
for ( it = v.begin(); it != v.end(); it++)
cout << " " << *it;
cout << endl;
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <bits/stdc++.h>
using namespace std;
int main()
{
char s[200],c;
int i,l,j;
while(gets(s))
{
l=strlen(s);
for(i=0;i<l;i++)
{
if(s[i]=='A' ||s[i]=='B' ||s[i]=='C' )
printf("2");
else if(s[i]=='D' ||s[i]=='E' ||s[i]=='F' )
printf("3");
else if(s[i]=='G'||s[i]=='H'||s[i]=='I')
printf("4");
else if(s[i]=='J'||s[i]=='K'||s[i]=='L')
printf("5");
else if(s[i]=='M'||s[i]=='N'||s[i]=='O')
printf("6");
else if(s[i]=='P'||s[i]=='Q'||s[i]=='R'||s[i]=='S')
printf("7");
else if(s[i]=='T'||s[i]=='U'||s[i]=='V')
printf("8");
else if(s[i]=='W'||s[i]=='X'||s[i]=='Y'||s[i]=='Z')
printf("9");
else
printf("%c",s[i]);
}
printf("\n");
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <bits/stdc++.h>
using namespace std;
char in[1002];
int main()
{
int j,i,count,sum,totat,digit,deg,l,q,d;
bool flag;
while(1)
{
flag=true;
sum=0;
l=strlen(gets(in));
if(l==1 && in[0]=='0')return 0;
for(i=0;i<l;i++)
sum+=in[i]-'0';
if(sum%9==0) flag=true;
else
flag=false;
count=1;q=0;d=sum;
if(flag==true)
while(d!=9 and d>9)
{
while(d!=0)
{
q+=d%10;
d=d/10;
}
d=q;
count++;
}
if(flag==true)
printf("%s is a multiple of 9 and has 9-degree %d.\n",in,count);
else if(flag==false)
printf("%s is not a multiple of 9.\n",in);
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <iostream>
#include <string.h>
using namespace std;
int main()
{
char s[150];
int i,j,total,result,l;
while(gets(s))
{
total=0;
l=strlen(s);
for(i=0;i<l;i++)
{
switch(s[i])
{
case 'a':
total=total+1; break;
case 'b':
total=total+2; break;
case 'c':
total=total+3; break;
case 'd':
total=total+4; break;
case 'e':
total=total+5; break;
case 'f':
total=total+6; break;
case 'g':
total=total+7; break;
case 'h':
total=total+8; break;
case 'i':
total=total+9; break;
case 'j':
total=total+10; break;
case 'k':
total=total+11; break;
case 'l':
total=total+12; break;
case 'm':
total=total+13; break;
case 'n':
total=total+14; break;
case 'o':
total=total+15; break;
case 'p':
total=total+16; break;
case 'q':
total=total+17; break;
case 'r':
total=total+18; break;
case 's':
total=total+19; break;
case 't':
total=total+20; break;
case 'u':
total=total+21; break;
case 'v':
total=total+22; break;
case 'w':
total=total+23; break;
case 'x':
total=total+24; break;
case 'y':
total=total+25; break;
case 'z':
total=total+26; break;
case 'A':
total=total+27; break;
case 'B':
total=total+28; break;
case 'C':
total=total+29; break;
case 'D':
total=total+30; break;
case 'E':
total=total+31; break;
case 'F':
total=total+32; break;
case 'G':
total=total+33; break;
case 'H':
total=total+34; break;
case 'I':
total=total+35; break;
case 'J':
total=total+36; break;
case 'K':
total=total+37; break;
case 'L':
total=total+38; break;
case 'M':
total=total+39; break;
case 'N':
total=total+40; break;
case 'O':
total=total+41; break;
case 'P':
total=total+42; break;
case 'Q':
total=total+43; break;
case 'R':
total=total+44; break;
case 'S':
total=total+45; break;
case 'T':
total=total+46; break;
case 'U':
total=total+47; break;
case 'V':
total=total+48; break;
case 'W':
total=total+49; break;
case 'X':
total=total+50; break;
case 'Y':
total=total+51; break;
case 'Z':
total=total+52; break;
}
}
result=1;
for(i = 2; i <=total/2; i++)
if(total % i == 0)
{
result=0; break;
}
if(result==1)
printf("It is a prime word.\n");
else
printf("It is not a prime word.\n");
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <bits/stdc++.h>
using namespace std;
int main()
{
char input[1002];
int i,j,l,sum;
while(1)
{
gets(input);
sum=0;
l=strlen(input);
if(input[0]=='0' and l==1) break;
for(i=0;i<l;i+=2)
{
sum+=input[i]-'0';
}
for(i=1;i<l;i+=2)
{
sum-=input[i]-'0';
}
if((sum) % 11) printf("%s is not a multiple of 11.\n",input);
else
printf("%s is a multiple of 11.\n",input);
}
return 0;
}
by Jesmin Akther | Jan 10, 2019 | Problem Solving, UVa
#include <stdio.h>
#include<iostream>
#include<math.h>
#define max 1000005
using namespace std;
bool flag[max];
int primes[max];
int cnt;
int sieve(int z)
{
int i,j;
for(i=2; i<=max; i++)
{
primes[i]=1;
}
for(i=2; i<=sqrt(max); i++)
{
for(j=2; i*j<=max; j++)
{
primes[i*j]=0;
}
}
}
int main()
{
int count,b,k,n;
sieve(max);
while(scanf("%d",&n) && n)
{
count=0;
for(k=2; k<n; k++)
{
b=n-k;
if(primes[b]&& primes[k])
{
count++;
break;
}
}
if(count>0)
{
printf("%d:\n",n);
printf("%d+%d\n",k,b);
}
else
{
printf("%d:\n",n);
printf("NO WAY!\n");
}
}
return 0;
}
