by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1110 Solution
#include<stdio.h>
#include<math.h>
#include<string.h>
#define fi(a, b) for(int i=a; i<b; i++)
#define fj(a, b) for(int j=a; j<b; j++)
#define fk(a, b) for(int k=a; k<b; k++)
#define sf scanf
#define pf printf
#define ssf sscanf
#define mem(name, value) memset(name, value, sizeof(name))
#define Max 1000
#define oo 1000000000
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef char C;
int dp[1001];
int X, Y, M;
int rec(int N)
{
if(N <= 1) return X % M;
if(dp[N] != -1) return dp[N];
return dp[N] = (rec(N / 2) % M * rec(N - N / 2) % M) % M;
}
int main()
{
int test_cases, cases = 1;
int N;
sf("%d %d %d", &N, &M, &Y);
int arr[1005], kount = 0;
mem(arr, -1);
int i = 0;
for(i = 0; i < M; i++)
{
mem(dp, -1);
X = i;
if(rec(N) == Y)
arr[kount++] = i;
}
if(kount == 0)
{
pf("-1\n");
return 0;
}
pf("%d", arr[0]);
for(i = 1; i < kount; i++)
pf(" %d", arr[i]);
puts("");
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1120 Solution
#include <stdio.h>
int main(void){
unsigned n,p,s;
scanf("%u",&n);
for(p=44720;n<(s=(p*(p+1))>>1)||(n-s)%p;p--);
printf("%u %u\n",1+(n-s)/p,p);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1196 Solution
#include <stdio.h>
int b[1000000];
int a[15000];
int main()
{
int i,j,k,n,m,c=0,first,last,mid;
scanf("%d\n",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
scanf("%d\n",&m);
for(i=0;i<m;i++)
{
scanf("%d",&b[i]);
}
first=0,last=n-1;
mid=(first+last)/2;
while(first<=last)
{
mid=(first+last)/2;
if(a[mid]<m)
{
first=mid+1;
}
else if(a[mid]==m)
{
printf("%d",mid+1);
break;
}
else
last=mid-1;}
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1197 Solution
#include <stdio.h>
#include <conio.h>
int main()
{
int X, Z,t;
char Y;
scanf("%d",&t);
while(t--){
scanf("%c%d",&Y, &Z);
switch ( Y )
{
case 'a': Y = 1;
break;
case 'b': Y = 2;
break;
case 'c': Y = 3;
break;
case 'd': Y = 4;
break;
case 'e': Y = 5;
break;
case 'f': Y = 6;
break;
case 'g': Y = 7;
break;
case 'h': Y = 8;
break;
default:
break;
}
if (Y == 1 && Z == 1 || Y == 1 && Z == 8 || Y == 8 && Z == 1 || Y == 8 && Z == 8) {
printf("2");
} else if(Y == 1 && Z == 2 || Y == 1 && Z == 7 || Y == 2 && Z == 1 || Y == 2 && Z == 8 || Y == 7 && Z == 1 || Y == 7 && Z == 8 || Y == 8 && Z == 2 || Y == 8 && Z == 7) {
printf("3");
} else if (Y == 1 && Z == 3 || Y == 1 && Z == 4 || Y == 1 && Z == 5 || Y == 1 && Z == 6 || Y == 8 && Z == 3 || Y == 8 && Z == 4 || Y == 8 && Z == 5 || Y == 8 && Z == 6 || Z == 1 && Y == 3 || Z == 1 && Y == 4 || Z == 1 && Y == 5 || Z == 1 && Y == 6 || Z == 8 && Y == 3 || Z == 8 && Y == 4 || Z == 8 && Y == 5 || Z == 8 && Y == 6 || Y == 2 && Z == 2 || Y == 2 && Z == 7 || Y == 7 && Z == 2 || Y == 7 && Z == 7) {
printf("4");
} else if(Y == 2 && Z == 3 || Y == 2 && Z == 4 || Y == 2 && Z == 5 || Y == 2 && Z == 6 || Y == 7 && Z == 3 || Y == 7 && Z == 4 || Y == 7 && Z == 5 || Y == 7 && Z == 6 || Z == 2 && Y == 3 || Z == 2 && Y == 4 || Z == 2 && Y == 5 || Z == 2 && Y == 6 || Z == 7 && Y == 3 || Z == 7 && Y == 4 || Z == 7 && Y == 5 || Z == 7 && Y == 6) {
printf("6");
} else if(Y == 3 && Z == 3 || Y == 3 && Z == 4 || Y == 3 && Z == 5 || Y == 3 && Z == 6 || Y == 4 && Z == 3 || Y == 4 && Z == 4 || Y == 4 && Z == 5 || Y == 4 && Z == 6 || Y == 5 && Z == 3 || Y == 5 && Z == 4 || Y == 5 && Z == 5 || Y == 5 && Z == 6 || Y == 6 && Z == 3 || Y == 6 && Z == 4 || Y == 6 && Z == 5 || Y == 6 && Z == 6) {
printf("8");
} }
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
#include<stdio.h>
#include<math.h>
unsigned long a[70000];
int main()
{
unsigned long i,k,n;
long double l;
scanf("%lu\n",&n);
for(i=0;i<n;i++)
{
scanf("%lu",&a[i]);
}
for(i=0;i<n;i++)
{
l=((sqrt(-7+8*a[i])-1)/ 2);
k=l;
if(l==k)
printf("1 ");
else
printf("0 ");
}
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1224 Solution
#include <stdio.h>
int main() {
long int n, m,o;
scanf("%ld%ld", &n, &m);
o=n <= m ? 2 * n - 2 : 2 * m - 1;
printf("%u",o);
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1225 Solution
#include <stdio.h>
int main() {
int n, m;
scanf("%d", &n);
m=2*n-2;
printf("%d",m);
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1234 Solution
#include <stdio.h>
int a[10001];
int main()
{
int ans=0,ans1=0,k,n,i;
scanf("%d %d",&n,&k);
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}
if(a[i]<k){
ans+=k-a[i];
}
else{
ans1+=a[i]-k;
}
printf("%d %d",ans,ans1);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
#include <stdio.h>
int main()
{
long long int a,b,l=0,r=0;
scanf("%lld",&a);
r=a%7;
printf("minr=%lld",r);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1263 Solution
#include <stdio.h>
double a[10004];
int main()
{
int i,m,n,l,o;
double d=0;
scanf("%d %d",&n,&m);
l=m;
while(m--)
{
scanf("%d",&o);
a[o]++;
}
for(i=0;i<n;i++)
{
d=(a[i]/l)*100;
printf("%2.2lf%%\n",d);
}
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
#include <stdio.h>
int main()
{
int m,n,l=0;
scanf("%d %d",&n,&m);
l=n*(m+1);
printf("%d",l);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1290 Solution.
#include <stdio.h>
#include <stdlib.h>
int a[25001];
int cmp (const void *a, const void *b)
{
return (*(int*)a - *(int*)b);
}
int main ()
{
int N,i;
scanf("%d",&N);
for(i=0;i<N;i++)
scanf("%d",&a[i]);
qsort(a,N,sizeof(int),cmp);
for(i=N-1;i>-1;i--)
printf("%d",a[i]);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1293 Solution
#include <stdio.h>
int main()
{
int n,a,b,sum;
scanf(“%d %d %d”,&n,&a,&b);
sum=2*n*a*b;
printf(“%d”,sum);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1313 Solution
#include<stdio.h>
int main()
{
int a[155],b[155],nn;
int n,i,j,count,h,k,m=0;
scanf("%d\n",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for (i=0;i<n;i++)
{
count=0;
for (j=0;j<n;j++)
{
if (a[i]==a[j])
count++;
}
if (count>=4)
{
h=0;
for(k=0;k<i;k++)
{
if(a[i]!=a[k])
h++;
}
if (h==i)
{
(b[m++]==a[i]);
}
}
}
nn=m;
printf("%d",nn);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
#include <stdio.h>
int main()
{
int a,b,res;
scanf("%d %d",&a,&b);
if(b%2==0)
{
printf("%d",((b-a)+1)/2);
}
else
{
printf("%d",((b-a)+2)/2);
}
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
#include <stdio.h>
int main()
{
int a,b,rest,test,i;
i=10;
scanf("%d %d",&a,&b);
i=a+b-1;
rest=i-a;
test=i-b;
printf("%d %d",rest,test);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
#include <stdio.h>
int main()
{
int n,a,b,c,d,e,f,g,h,i,j,a1,b1,c1,d1,e1,f1,g1,h1,i1,j1,first,last,first1,last1,sum1,sum2;
scanf("%d",&n);
sum1=n+1;
sum2=n-1;
a=sum1/100000;
b=sum1%100000;
c=b/10000;
d=b%10000;
e=d/1000;
f=d%1000;
g=f/100;
h=f%100;
i=h/10;
j=h%10;
first=a+c+e;
last=g+i+j;
a1=sum1/100000;
b1=sum2%100000;
c1=b1/10000;
d1=b1%10000;
e1=d1/1000;
f1=d1%1000;
g1=f1/100;
h1=f1%100;
i1=h1/10;
j1=h1%10;
first1=a1+c1+e1;
last1=g1+i1+j1;
if((first1==last1)||(first==last)){
printf("Yes");
}
else{
printf("No");
}
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1567 Solution
#include <stdio.h>
int main()
{
char c;
int i,sum=0;
while(scanf("%c",&c)==1){
if(c=='a')
sum+=1;
else if(c=='b')
sum+=2;
else if(c=='c')
sum+=3;
else if(c=='d')
sum+=1;
else if(c=='e')
sum+=2;
else if(c=='f')
sum+=3;
else if(c=='g')
sum+=1;
else if(c=='h')
sum+=2;
else if(c=='i')
sum+=3;
else if(c=='j')
sum+=1;
else if(c=='k')
sum+=2;
else if(c=='l')
sum+=3;
else if(c=='m')
sum+=1;
else if(c=='n')
sum+=2;
else if(c=='o')
sum+=3;
else if(c=='p')
sum+=1;
else if(c=='q')
sum+=2;
else if(c=='r')
sum+=3;
else if(c=='s')
sum+=1;
else if(c=='t')
sum+=2;
else if(c=='u')
sum+=3;
else if(c=='v')
sum+=1;
else if(c=='w')
sum+=2;
else if(c=='x')
sum+=3;
else if(c=='y')
sum+=1;
else if(c=='z')
sum+=2;
else if(c=='.')
sum+=1;
else if(c==',')
sum+=2;
else if(c=='!')
sum+=3;
else if(c==' ')
sum+=1;
}
printf("%d",sum);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1581 Solution
#include <stdio.h>
int a[1001];
int main()
{
int i,j=1,n;
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%d",&a[i]);
}
if(a[i]==a[j]){
i++;
j++;
}
printf("%d %d",i,j);
return 0;
}
by Jesmin Akther | Dec 24, 2018 | Timus Problem Solution
Timus Problem 1585 Solution
#include<stdio.h>
#include<string.h>
int main()
{
char a[1001][30];
char b[2][30];
int n,i,j,count,k,x=0,l=0;
scanf("%d\n",&n);
for(i=0;i<n;i++)
{
gets(a[i]);
}
for(j=0;j<n;j++)
{
count=0;
for(k=0;k<n;k++)
{
if(strcmp(a[j],a[k])==0)
{
count++;
}
}
if(count>x)
{
strcpy(b[l],a[j]);
}
x=count;
}
puts(b[l]);
return 0;
}
