#include<bits/stdc++.h> using namespace std; int main() { int tc,a[13]; scanf ("%d", &tc); printf ("Lumberjacks:\n"); while ( tc-- ) { for ( int i = 0; i < 10; i++ ) { scanf ("%d", &a [i]); } bool flag = true; if (a [0] > a [1]) flag =...
#include<bits/stdc++.h> using namespace std; int main() { int tc,a[13]; scanf ("%d", &tc); printf ("Lumberjacks:\n"); while ( tc-- ) { for ( int i = 0; i < 10; i++ ) { scanf ("%d", &a [i]); } bool flag = true; if (a [0] > a [1]) flag =...
#include<bits/stdc++.h> using namespace std; int main() { int t; double d, sum; int j=0; scanf("%d",&t); while(t--) { sum = 0; for(int i=0; i<12; i++) { scanf("%lf",&d); sum = sum + d; } double avg = sum/12.00; printf("%d $",++j);...
#include<bits/stdc++.h> using namespace std; int main() { long int N,X,r; int T,i,j=1; cin>>T; while(T--) { cin>>N; printf("Case %d:",j++); r=sqrt(N); for(i=r-1;i>0;i--) { X=N-i*i; if(X%i==0) cout<<" "<<X; }...
#include<bits/stdc++.h> using namespace std; int main() { double F,C,c,d; int tc,i; while(cin>>tc) { for(i=1;i<=tc;i++) { cin>>c>>d; F = 9*c/5+d; C = F*5/9; printf("Case %d: %.2lf\n",i,C); } } return 0; }...
#include <bits/stdc++.h> #include <vector> using namespace std; char letters[100001]; int main() { int i; while ( scanf ("%s", &letters) != EOF ) { int l = strlen (letters); list <char>beiju; list <char>::iterator it =...
#include <cstdio> #include <vector> using namespace std; int main() { int n, m, N,V, k; vector<vector<int> > v; while (scanf("%d %d", &n, &m) != EOF) { v.assign(1000000, vector<int>()); for (int i = 1; i <= n;...
#include <bits/stdc++.h> using namespace std; int main() { int a,i,n,j,s; while(scanf("%d",&a)==1 && a>0) { s=0; for(i=1;i<=a;i++) s+=(i*i); cout<<s<<endl; } return 0; }...
#include <bits/stdc++.h> using namespace std; int main() { int H,W,L,tc,c=1; cin>>tc; while(tc--) { cin>>L>>W>>H; printf("Case %d: ",c++); if(L<=20 and W<=20 and H<=20) cout<<"good"<<endl; else...
#include<stdio.h> int main() { int n,m,ro,a[20],num,i,j,digit,k,count; while(scanf("%d %d",&n,&m)==2) { ro=0; for(i=n;i<=m;i++) { num=i; digit=0; while(num>0) { a[digit++]=num%10; num/=10; } count=0; for(j=0;j<digit-1;j++) {...
#include <bits/stdc++.h> using namespace std; long int a[5],b[5]; int main() { bool flag; while(cin>>a[0]>>a[1]>>a[2]>>a[3]>>a[4]>>b[0]>>b[1]>>b[2]>>b[3]>>b[4]) { flag=false; for(int...
#include <bits/stdc++.h> using namespace std; int main() { int num,tc,sum,j; vector<int>v; cin>>tc; while(tc--) { cin>>num; v.push_back(num); int tempo=num; int i=0; while(tempo) { v[i]=tempo%10; tempo/=10; i++; } for(sum = j...
#include <bits/stdc++.h> using namespace std; int main() { int h1,m1,h2,m2,hr,mr,tr; while(cin>>h1>>m1>>h2>>m2) { if(h1==0 && m1==0 && h2==0 && m2==0) break; hr=h2-h1; mr=m2-m1; tr=(hr*60)+mr;...
#include <bits/stdc++.h> using namespace std; int main() { double d,u,v,t1,t2; int count,n; cin>>n; count=1; while(n--) { cin>>d>>v>>u; if(u==0 ||v==0 ||u<=v) { printf("Case %d: can't determine\n",count); continue; } t1=d/u;...
//UVA Problem 11462 Solution /** * @Author : Jesmin Akther */ #include <stdio.h> long int n, array[2000005], c, d, t; int main() { while(scanf("%ld", &n)==1 ) { if(n==0) break; for (c = 0; c < n; c++) { scanf("%ld", &array[c]); } for (c = 1 ; c <=...
Here we will discuss pros and cons of all the questions of the passage with step by step Solution included Tips and Strategies. Reading Passage 1 –Roman Tunnels IELTS Cambridge 16, Test 4, Academic Reading Module, Reading Passage 1 Questions 1-6. Label the diagrams...
Reading Passage 1: Roman Shipbuilding and Navigation, Solution with Answer Key , Reading Passage 1: Roman Shipbuilding and Navigation IELTS Cambridge 16, Test 3, Academic Reading Module Cambridge IELTS 16, Test 3: Reading Passage 1 – Roman Shipbuilding and...
Cambridge IELTS 16, Test 1, Reading Passage 1: Why We Need to Protect Bolar Bears, Solution with Answer Key Cambridge IELTS 16, Test 1: Reading Passage 1 – Why We Need to Protect Bolar Bears with Answer Key. Here we will discuss pros and cons of...
PASSAGE 1: THE RETURN OF THE HUARANGO QUESTIONS 1-5: COMPLETE THE NOTES BELOW. 1. Answer: water Key words: access, deep, surface Paragraph 2 provides information on the role of the huarango tree: “it could reach deep water sources”. So...