Timus Problem 1001(Reverse Root) : Reverse Root is a mathematical problem. To read details : http://acm.timus.ru/problem.aspx?space=1&num=1001 Proplem description: This is a straightforward problem.Just do square root. Problem Solution: #include...

Timus Problem 1005 (Stone pile): This problem is for beginner in timus online judge. Problem details link : http://acm.timus.ru/problem.aspx?space=1&num=1005 Problem Description: just find the minimal difference among a value range. Usage Tips: >> Right...

Problem details link: http://acm.timus.ru/problem.aspx?space=1&num=1009 Problem Description: This is a straightforward problem check validity of K-based numbers. Problem Solution: #include <stdio.h> int main() { int m[1009],n,k; scanf("%d...

Timus problem 1025 Democracy in Danger : This problem is for beginner. Problem Details: http://acm.timus.ru/problem.aspx?space=1&num=1025 Problem Description: This is a Straightforward problem. Timus Problem 1025 Solution #include<stdio.h> int main() { int...

Problem 1064 Binary Search details description : http://acm.timus.ru/problem.aspx?space=1&num=1064 Timus Problem 1064 Solution : #include <stdio.h> int main() { int n = 1; while(n <= 100) { printf("%d\n", n); n++; if(n > 10) { break; } n = 200; while(n...

Timus Problem 1068 Solution #include <stdio.h> #include <stdlib.h> long int sum[10000]; int a[10000]; int main() { int i,n; sum[0]=1; scanf("%d",&n); for(i=0;i<=n;i++){ scanf("%d",&a[i]);} for(i=0;i<=n;i++){ sum[i]+=a[i];...

Timus Problem 1083 Solution #include <stdio.h> int main() { char ch; int n,k=0,f=1; scanf("%d %c",&n,&ch); if(ch=='!'){ printf("%c",ch); k++;} else if(ch=='\n') printf("%c",ch);{ break;} for(n=1;n<=10;n++) f*=f*(n-2k); if(f<=0){ break; }...

Timus Problem 1086 Solution #include <iostream> #include <stdio.h> using namespace std; bool flag[20000005]; int primes[20000005]; 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;...

#include <stdio.h> int main() { int i,swap,j,n,l,sum=0,k,ara[100][100]; scanf("%d",&n); for(i=0;i<n;i++){ scanf("%d %d",&ara[i][j]); } for(i=0;i<n;i++) { for(j=0;j<n-i;j++) { if(ara[j]<ara[j+1]) { ara[i]=ara[i+1]; ara[i+1]=ara[j];...

#include <stdio.h> #include <ctype.h> int main() { long num=0,k,v,max=1; char ch; ch=getchar(); while(ch!=EOF) { if(isdigit(ch)) v=ch-'0'; else v=ch-'A'+10; num+=v; if(v>max) max=v; ch=getchar(); } for(k=max;k<36;k++) if(num%k==0) break; if(k<36)...

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...

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; }  

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]); }...

  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;...

#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 ");...

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); }  

Timus Problem 1225 Solution #include <stdio.h> int main() { int n, m; scanf("%d", &n); m=2*n-2; printf("%d",m); }  

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;...

#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; }  

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; }...