Devika is addicted to meat! malik wants to keep
#include<iostream> using namespace std;
#define f(n) for(n=n;n>0;–n) int main()
{
int n,r=0,m=100,x,y;
cin>>n; f(n){
cin>>x>>y; if(y<m) m=y; r+=m*x;
}
printf("%d",r);
}
Students in a class are making towers of blocks.
#include<iostream>
using namespace std;
int m, n, x; int main()
{
cin>>n>>m; x = 1;
while((x/2) < n || (x/3) < m || x/2 + x/3 – x/6 < m+n)
++x;
cout << x; return 0;
}
shiv has given a rebus of form ?+?
#include <bits/stdc++.h> using namespace std; int p = 1, n, j, a[105]; char c;
int main()
{
a[j++] = 1;
while (cin>>c && c != '=')
{
if (c == '-') p–, a[j++] = -1; if (c == '+') p++, a[j++] = 1;
}
cin>>n;
for(int i=0;i<j;i++)
{
if(a[i]>0)while (p<n && a[i]<n) a[i]++, p++;
else while (p>n&&a[i]<0 && a[i]>-n) a[i]–, p–;
}
if (p != n) { cout << "Impossible\n"; return 0; } cout << "Possible\n";
for(int i=0;i<j;i++)
cout << (i ? (a[i]<0 ? "- " : "+ ") : "") << abs(a[i]) << " ";
cout << "= " << n;
}
It's a very unfortunate day for lavanya today
#include <bits/stdc++.h> using namespace std;
#define res cin>>a[i],num+=a[i]; #define f1 for(int i=1;i<=n;i++) double n,v,a[25],b[25],sum,mx=1e9; int main(){
cin>>n>>v; f1{
cin>>a[i]; sum+=a[i];
}
for(int i=1;i<=n;i++)
cin>>b[i]; for(int i=1;i<=n;i++)
mx=min(mx,b[i]/a[i]);
cout << fixed<<setprecision(1)<<min(mx*sum,v); return 0;
}
A stealing got into a matches warehouse and wants to steal
#include<bits/stdc++.h> using namespace std;
#define res cin>>a>>b; cin>>s>>d; int n,m,s,a,b,d[11];
int main(){ cin>>n>>m;
while(m–)cin>>a>>b,d[b]+=a;
for(int i=10;i>0&&n>0;i–)s+=i*min(n,d[i]),n-=d[i]; cout<<s;
}
A long time ago, in a galaxy far far away
#include<bits/stdc++.h>
int a,i;
int main()
{std::string s,t; std::cin>>s>>t;
for(i=s.find(t);i+1;++a,i=s.find(t,i+t.size()));std::cout<<a;}
the spring is coming and it means that a lot
#include<bits/stdc++.h> using namespace std; map <string,int> p;
int n,m,g[102],c[102],cnt; string s;
int main()
{
cin>>n>>m; for(int i=0;i<n;i++)
cin>>g[i]; sort(g,g+n);
for(int i=0;i<m;i++){
cin>>s; if(!p[s])p[s]=++cnt; c[p[s]]++;
}
sort(c+1,c+cnt+1); int num=0;
for(int i=1;i<=cnt;i++)
num+=c[i]*g[cnt-i]; cout<<num<<" ";
num=0;
for(int i=1;i<=cnt;i++)
num+=c[i]*g[n-cnt+i-1]; cout<<num;
return 0;
}
If There are n banks in the city where vishnu lives
#include<bits/stdc++.h> using namespace std; #define maxs long long
map <maxs,maxs> a; maxs i,n,k,x,p;
int main(){ cin>>n;
for(;i<n;i++)cin>>x,k+=x,a[k]++,p=max(p,a[k]); cout<<n-p;
}
You have infinite cards for each number between 1 and N
#include <bits/stdc++.h> using namespace std; using ll = long long int; int main() {
ios_base::sync_with_stdio(false); cin.tie(NULL);
cout.tie(NULL);
//preSum(); ll t;
cin>>t; while(t–){
ll n; cin>>n; if(n==1)
printf("1\n"); else if(n==2)
printf("4\n"); else if(n==3)
printf("10\n");
else
printf("%lld\n",9*n-18);
}
}
Samy has bought a box of chocolate and has brought them to Anand house.
#include<stdio.h>
int function(int arr[],int i,int j,int memo[][1001],int k)
{
if(i>j)
return 0; if(arr[i]!=arr[j])
return 0; if(i==j)
return 1; if(memo[i][j]!=-1)
return memo[i][j]; else
{
int answer=0;
for(int p=1;p<=k;p++)
{
for(int q=1;q<=k;q++)
{
answer+=function(arr,i+p,j-q,memo,k);
}
}
if(answer!=0) answer=1; memo[i][j]=answer; return answer;
}
}
int main()
{
int n,k; scanf("%d%d",&n,&k); int j,arr[n+1]; for(j=1;j<=n;j++)
scanf("%d",&arr[j]); int memo[1001][1001];
// int answer=0; int i;
for(i=0;i<=1000;i++)
{
for(j=0;j<=1000;j++)
{
memo[i][j]=-1;
}
}
int answer=function(arr,1,n,memo,k); if(answer==0)
printf("NO\n"); else
printf("YES\n");
}
There are N knights sitting at the round table at an equal distance from each other.
#include<bits/stdc++.h> using namespace std; int n,a[100020],z;
int main()
{
cin>>n;
for(int i=0;i<n;i++) cin>>a[i]; for(int i=1;i<=n/3;i++)
if(n%i==0)
for(int j=0;j<i;j++)
{
z=1;
for(int k=j;k<n;k+=i) z&=a[k]; if(z)
{cout<<"YES";return 0;}
}
cout<<"NO"; return 0;
}
Venkat plays the age of emperor II. He was bored of playing
#include <bits/stdc++.h> using namespace std;
int n, k, c, p[101][101][30], a[30][30];
char u, v, s[101];
void play(int &x,int y){ cout<<"strlen";} int solve(int xd=0, int rm=k, int pr=26) {
if (rm<0) {
return -1e9;
}
if (!s[xd]) {
return 0;
}
int& rt=p[xd][rm][pr]; if (~rt) {
return rt;
}
rt=solve(xd+1, rm, s[xd]-'a')+a[pr][s[xd]-'a'];
for (int i=0; i<26; i++) {
rt=max(rt, solve(xd+1, rm-1, i)+a[pr][i]);
}
return rt;
}
int main() {
cin>>s>>k>>n; while (n–) {
cin>>u>>v>>c; a[u-'a'][v-'a']=c;
}
memset(p, -1, sizeof(p)); cout << solve();
}
This is the easy version of the problem. The only difference is maximum value
#include<bits/stdc++.h> #define int long long using namespace std;
int const M=5000000;int i,j,n,s,x,e[M+100],f[M+100],d[M+100]; signed main(){
cin>>n;
for (i=1;i<=n;i++) scanf("%lld",&x),f[x]++; for (i=1;i<=M;i++)
for (j=i;j<=M;j+=i)
e[i]+=f[j];
for (i=M;i>0;i–){
for (s=0,j=i*2;j<=M;j+=i) s=max(s,d[j]-e[j]*i);
d[i]=e[i]*i+s;
}
printf("%lld\n",d[1]); return 0;
}
There are N sprinklers in a field. Each sprinkler has some range up to where it can sprinkle
#include<bits/stdc++.h> using namespace std; #define mod 1000000007 #define endl "\n"
#define test ll t; cin>>t; while(t–) typedef long long int ll;
int main() { test
{
ll n,q; cin>>n>>q; vector<ll>x(n),r(n); for(auto &it:x) cin>>it; for(auto &it:r) cin>>it; vector<ll>ans(4*n+5,0); for(int i=0;i<n;i++){
ll left=x[i]-r[i]+2*n;
ll right=x[i]+r[i]+2*n+1; if(x[i]>0){
left=max(left,2*n);
}
else{
right=min(right,2*n);
}
ans[left]++; ans[right]–;
}
for(int i=1;i<4*n+5;i++){ ans[i]+=ans[i-1];
}
while(q–){
int inp; cin>>inp; inp+=2*n; cout<<ans[inp]<<endl;
}
}
return 0;
}
Professor Wiki has performed some experiments on rays. The setup for n rays
#include<bits/stdc++.h> using namespace std; int n,x,i;
int a[1000020]; int p[1000020]; int f[1000020];
int main()
{
cin>>n;
for(i=0;i<n;i++)
{
cin>>x; p[x]=i;
}
for(i=0;i<n;i++)
{
scanf("%d",&x);
a[i]=-p[x]-1;
}
for(i=0;i<n;i++)
*lower_bound(f,f+n,a[i])=a[i];
int zero=0;
printf("%ld\n",lower_bound(f,f+n,zero)-f); return 0;
}
There are N sprinklers in a field. Each sprinkler has some range up
#include <stdio.h> int min(int a,int b){
if (a<b){
return(a);
}
else{
return(b);
}
}
int max(int a,int b){
if (a>b){
return(a);
}
else{
return(b);
}
}
int main(){
int t; scanf("%d",&t); for(int i=0;i<t;i++){
int n,q,k;
scanf("%d %d",&n,&q);
int a[n],b[n]; for(int j=0;j<n;j++){
scanf("%d",&a[j]);
}
for(int m=0;m<n;m++){ scanf("%d",&b[m]);
}
int x=(4*n)+1; int sum[x];
for (int m=0;m<x;m++){
sum[m]=0;
}
for (int m=0;m<n;m++)
{
int low=a[m]-b[m]; int up=a[m]+b[m]+1; if (a[m]>0){
low=max(1,low);
}
if (a[m]<0){
up=min(0,up);
}
low +=2*n; up +=2*n; sum[low]+=1; sum[up]-=1;
}
for (int y=1;y< 4*n+1;y++){
sum[y]+=sum[y-1];
}
for(int m=0;m<q;m++){
scanf("%d",&k); printf("%d\n",sum[k+(2*n)]);
}
}
}
Bob goes to the fruit shop to buy apples. There are N apples numbered from 1 to N
#include<bits/stdc++.h> using namespace std;
int i,n, m, sum, a[1002][2]; void sol()
{
cin >> n >> m;
for(int i = 1; i <= m; i ++)a[i][0] = a[i][1] = -1;
a[0][0] = 0;
a[0][1] = -1;
sum = 0; for(i=1;i<=n;i++)
{
int v, p;
cin >> v >> p;
for(int j = min(m-p/2, sum); j >= 0; j –)
{
if(a[j][1] != -1 && j + p <= m)a[j+p][1] = max(a[j+p][1], a[j][1] + v); if(a[j][0] != -1)
{
if(j + p <= m)a[j+p][0] = max(a[j+p][0], a[j][0] + v);
a[j+p/2][1] = max(a[j+p/2][1], a[j][0] + v);
}
}
sum = min(m, sum + p);
}
int ans =0 ;
for(int i = 1; i <= m; i ++)ans = max(ans, max(a[i][0], a[i][1])); cout << ans << '\n';
}
int main()
{
int ntest = 1; cin >> ntest;
while(ntest — > 0)sol();
}
Krishnes has given a directed acyclic graph with N vertices and M edges.
#include <bits/stdc++.h> using namespace std;
int T,n,m,pr[150010],l[150010],f[150010][2],a[4],b[4];
vector<int>E[150010]; void direction(int x,int c){} void pairs();
void pairs(){
scanf("%d%d",&n,&m);
for(int i=0;i<=n+1;i++) E[i].clear(),pr[i]=0,f[i][0]=f[i][1]=0; for(int i=1,u,v;i<=m;i++){
scanf("%d%d",&u,&v); if(u+1==v) pr[v]=1; else E[v].push_back(u);
}
pr[1]=pr[n+1]=1;
for(int i=2;i<=n;i++) E[i].push_back(0),E[n+1].push_back(i-1); int L=0,R=n+1;
while(L<=n&&pr[L+1]) L++; while(R&&pr[R]) R–;
if(R==0) return printf("%lld\n",1ll*n*(n-1)/2),void(); for(int i=1;i<=n;i++) l[i]=pr[i]?l[i-1]:i;
f[L][0]=1,f[L][1]=2;
for(int i=L;i<=n;i++) for(int u:E[i+1]) for(int k=0;k<2;k++) if(l[i]<=u+1) f[i][k]|=f[u][k^1]; for(int i=L;i>=1;i–) for(int u:E[i+1]) for(int k=0;k<2;k++) if(l[i]<=u+1) f[u][k]|=f[i][k^1]; for(int i=0;i<4;i++) a[i]=b[i]=0;
for(int i=0;i<=L;i++) a[f[i][0]]++;
for(int i=R-1;i<=n;i++) b[f[i][0]]++; long long ans=0;
for(int p=0;p<4;p++) for(int q=0;q<4;q++) if(p&q) ans+=1ll*a[p]*b[q]; printf("%lld\n",ans-(L+1==R));
}
int main(){
scanf("%d",&T);
while(T–) pairs();
}
SESSION 6:
Given a chess board having NxN cells, you need to place N queens on the board in such
#include<iostream> using namespace std;
int n;
bool grid[10][10];
bool isSafe(int row, int col){ int i,j;
for(i=0;i<row;++i) if(grid[i][col]) return false; for(i=row,j=col;i>=0 and j>=0;–i,–j) if(grid[i][j]) return false; for(i=row,j=col;i>=0 and j<n;–i,++j) if(grid[i][j]) return false; return true;
}
bool solveQueen(int row){ if(row>=n) return true; for(int col=0;col<n;++col){ if(isSafe(row,col)){
grid[row][col]=true; if(solveQueen(row+1)) return true; grid[row][col]=false;
}
}
return false;
}
int main(){ cin>>n; int i;
if(solveQueen(0)){ for(i=0;i<n;++i){
for(int j=0;j<n;++j) cout<<grid[i][j]<<" "; cout<<endl;
}
}
else cout<<"Not possible\n"; return 0;
}
Chef started watching a movie that runs for a total of XX minutes.
#include<bits/stdc++.h> using namespace std;
void solve()
{
int x,y; cin>>x>>y;
cout<<(x-y)+y/2<<"\n";
}
int main()
{
int t;
t=1;
while(t–)
solve();
}
Samu is playing a shooting game in play station.
#include<bits/stdc++.h> using namespace std; #define MAXN 1010
#define MAXW 1010 double DP[MAXN][MAXW];
int main(){
int t; cin>>t; while(t–){
int X,Y,N,W,P1,P2; cin>>X>>Y>>N>>W>>P1>>P2;
for(int i=0;i<=N;i++){
for(int j=0;j<=W;j++){
DP[i][j]=0;
}
}
double p1 = 0.01 * P1; double p2 = 0.01 * P2; for (int i=0;i<=N;i++) DP[i][0] = 1;
for (int i=1;i<=W;i++)
DP[0][i] = 0;
for (int i=1;i<=N;i++){
for (int j=1;j<=W;j++) {
DP[i][j] = max(p1*DP[i-1][max(j-X,0)] + (1-p1)*DP[i-1][max(j,0)], p2*DP[i-1][max(j- Y,0)] + (1-p2)*DP[i-1][max(j,0)]);
}
}
printf("%.6f\n",DP[N][W]*100);
}
}
Given integer N, you need to find four integers A,B,C,D, such that they're all factors
#include<bits/stdc++.h> using namespace std; void solve(){}
int main(){
int t; cin>>t; while(t–){ long long n; cin>>n;
if(n&1 or n<4) cout<<"-1\n";
else if(!(n%4)) cout<<((n>>2)*(n>>2)*(n>>2)*(n>>2))<<'\n';
else if(!(n%6)) cout<<((n/6)*(n/6)*(n/3)*(n/3))<<'\n';
else if(!(n%10)) cout<<((n/10)*(n/5)*(n/5)*(n>>1))<<'\n'; else cout<<"-1\n";
}
return 0;
}
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There is a mysterious temple in mysteryland. The door of the temple is always
#include <bits/stdc++.h>
using namespace std;
int arr [100 + 5]; bool flag;
bool vis [100 + 5][10000 + 5];
int sum , n;
void rec(int index, int tot){
if(index == n){ // base case
if(tot == sum / 2) flag = true; //valid case return;
}
if(vis[index][tot]) return; // check if this state visited before
rec(index + 1, tot + arr[index]); // add this item to first box rec(index + 1, tot); // add this item to second box vis[index][tot] = true; // mark this state is visited
}
int main()
{
int t;
cin >> t;
while(t–){
cin >> n;
sum = 0; flag = false;
memset(vis, false, sizeof vis);
for(int i = 0; i < n; i++){
cin >> arr[i]; sum += arr[i];
}
if(sum % 2 != 0) cout << "NO\n"; // invalid case else {
rec(0 , 0);
if(flag) cout << "YES\n"; else cout << "NO\n";
}
}
return 0;
}
You are given three arrays a1…n,b1…n,c1…n and two numbers M and K.
#include <iostream> #include<bits/stdc++.h>
#define f1 for(i=0;i<n;i++) using namespace std;
long long int min(long long int x, long long int y){ if(x < y)
return x; else
return y;
}
int main(){
int n, m, K;
cin >> n >> m >> K;
long long int a[n],b[n],c[n]; long long int i,j,l;
int p,T = 0;
for(i = 0; i < n; i++)
cin >> a[i] >> b[i] >> c[i]; long long int lx,rx,ly,ry,lz,rz; cin >> lx >> rx;
cin >> ly >> ry; cin >> lz >> rz;
for(i = lx; i < min(rx, lx + m); i++){
for(j = ly; j < min(ry,ly + m);j++){
for(l = lz;l < min(rz,lz + m);l++){ T=0;
for(p = 0; p < n; p++){
if((a[p] * i + b[p] * j- c[p] * l) % m == 0) T++;
} if(T==K)
break;
}
if(l < min(rz,lz + m)) break;
}
if(j < min(ry,ly + m)) break;
}
if(i < min(rx, lx + m)){
cout << i << " " << j << " " << l;
}
else
cout << "-1" << endl;
}
The end semester exams are now approaching. Ritu is the computer
#include<iostream> #include<bits/stdc++.h> using namespace std; int arr[1001][1001];
int dp[1001][1001];
const int mod = 1e9+7; void solve(){
cout<<"for(i=0;i<N;i++)";}
long int decrease_path(int i , int j,int n)
{
if(arr[i][j]==1) return 1; if(dp[i][j]!=-1) return dp[i][j];
long int a = (i>0 && arr[i-1][j]<arr[i][j])?decrease_path(i-1,j,n):0;
long int b = (j>0 && arr[i][j-1]<arr[i][j])?decrease_path(i,j-1,n):0;
long int c = (i<n-1 && arr[i+1][j]<arr[i][j])?decrease_path(i+1,j,n):0; long int d = (j<n-1 && arr[i][j+1]<arr[i][j])?decrease_path(i,j+1,n):0; dp[i][j] = a+b+c+d+1;
dp[i][j]%=mod;
return dp[i][j];
}
int main()
{
cin.tie(NULL); ios_base::sync_with_stdio(false); int n;
cin>>n;
for(int i = 0;i<n;i++)
{
for(int j = 0;j<n;j++)
{
cin>>arr[i][j];
}
}
memset(dp,-1,sizeof(dp)); long int count = 0;
for(int i = 0;i<n;i++)
{
for(int j = 0;j<n;j++)
{
long int a = decrease_path(i,j,n); count+=a;
count%=mod;
}
}
cout<<count<<'\n';
}
You are given two numbers n and k. for each number in the interval [1,n], your task is to
#include<bits/stdc++.h> using namespace std;
using ll = long long;
long long f(int n, int k) { if (n == 0) return 0; long long res = (n/k);
return f(n/k, k) + n * (ll)(n+1) / 2ll – (res * (res + 1) / 2ll) * k;
}
int main () { int T = 1;
scanf("%d", &T);
assert(T >= 1 && T <= 300000); while(T–) {
int n, k;
scanf("%d%d", &n, &k); assert(n <= 1e9);
assert(k >= 2 && k <= 1e9);
printf("%lld\n", f(n, k));
}
return 0;
}
Lucky numbers are defined as the numbers consisting only of digits 3 and 5.
#include<bits/stdc++.h> using namespace std;
string solve(string& s)
{
int n = s.size(), i = 0;
while(i < n && (s[i] == '3' || s[i] == '5'))
++i;
if(i < n && (s[i] < '5'))
{
if(s[i] == '4')
s[i] = '5';
else
s[i]='3';
++i;
while(i<n)
{
s[i] = '3';
++i;
}
}
else
{
while(i >= 0 && (s[i] != '3'))
–i;
if(i < 0)
return string(n + 1, '3');
s[i] = '5';
++i;
while(i < n)
{
s[i] = '3';
++i;
}
}
return s;
}
int main()
{
ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
int t;
cin >> t; while(t–)
{
string s; cin >> s;
cout << solve(s) << endl;
}
}
SESSION 7: GRAPH COLOURING:-
Adiththi likes drawing. She has drawn many graphs already, both directed and not.
#include<bits/stdc++.h> using namespace std;
#define ggg int find(int p) unite(int p,int q) cin>>u>>v; int i,j,x,y,n,m,ans,f[502],a[502];
int find(int x){return (f[x]==x)?x:f[x]=find(f[x]);} int main (){
scanf("%d %d",&n,&m);
if (m>n) return printf("NO\n"),0; for (i=1;i<=n;i++) f[i]=i;
for (i=1;i<=m;i++){
scanf("%d %d",&x,&y);a[x]++;a[y]++;
if (find(x)==find(y)&&i!=n) return printf("NO\n"),0; f[find(x)]=find(y);
}
for (i=1;i<=n;i++)
if (a[i]>2)
return printf("NO\n"),0; printf("YES\n%d\n",n-m);
for (i=1;i<=n;i++)
while (a[i]<2){
ans=i+(n!=1);
for (j=ans;j<=n;j++)
if (a[j]<2&&(n<=2||m+1==n||find(i)!=find(j)))
{printf("%d
%d\n",i,j);m++;a[i]++;a[j]++;f[find(i)]=find(j);break;}
}
return 0;
}
Nowadays the one-way traffic is introduced all over the world in order to improve
#include<bits/stdc++.h> using namespace std; int s[105],e[105];
int main(){
int n,ans=0,res=0;cin>>n; while(n–){
int a,b,c;cin>>a>>b>>c; if(s[a]||e[b])res+=c,s[b]=e[a]=1; else s[a]=e[b]=1;
ans+=c;
}
cout<<min(res,ans-res);
}
Bragadesh got a job as a system administrator in X corporation.
#include<bits/stdc++.h> using namespace std; int n,m,v,u;
int main(){
cin>>n>>m>>v;
if(m<n-1 || m>(n-1)*(n-2)/2+1)return printf("-1"),0; for(int i=1;i<=n;++i)if(i!=v)printf("%d %d\n",i,v),u=i; m-=n-1;
if(m)for(int i=1;i<=n;++i)for(int j=i+1;j<=n;++j)if(i!=v && j!=u && i!=u && j!=v){
printf("%d %d\n",i,j); m–;
if(!m)return 0;
}
}
We call two numbers x and y similar if they have the same parity
#include<bits/stdc++.h>
#define solve int t,n,q,i,j,w,a[55],b[55]; while(t!=0) { int t,n,a[52],s,o,i;
int main(){
for(scanf("%d",&t);t–;){
for(scanf("%d",&n),o=0;i<n;) scanf("%d",a+i), o+=a[i++]&1;
for(std::sort(a,a+n),s=0;–i;) s|=(a[i]-a[i-1]==1); puts(o&1&(!s)?"NO":"YES");
}
}
One Egyptian boy called Aabid wants to present a string of beads to his friend
#include <bits/stdc++.h> using namespace std;
#define solve int beads(int len,int lim1,int lim2) cin>>n>>m; #define rep(i,s,t) for(I i=s;i<=t;++i)
#define c(f) memset(f,-1,sizeof f); #define R return
#define I int
#define L long long
L f[55][2][2],K,d;I a[55],n; L C(I l,I r,I x,I y){
if (l>r)R 1;L&F=f[l][x][y];if(~F)R F;F=0;
rep(i,0,1)rep(j,0,1)if(a[l]-!i&&a[r]-!j&&(l<r||i==j)&&(x||i<=j)&&(y||i<=(!j)))F+=C(l+1,r- 1,x||(i<j),y||(i<!j));R F;
}
I main(){
cin>>n>>K;c(a)c(f)if(C(1,n,a[1]=0,0)<++K)R cout<<-1,0;
rep(i,2,n){c(f)d=C(1,n,a[i]=0,0);K-=(a[i]=(d<K))*d;} rep(i,1,n)cout<<a[i];
}
Students of Winter Informatics School are going to live in a set of houses connected by underground passages.
#include "bits/stdc++.h" using namespace std;
#define mandatory cout<<"scanf("%d", &T); scanf("%d%d", &n, &m); while(T–)" #define pb push_back
#define mp make_pair
const int MX = 3e6 + 5; void solve2(){}
int N, M; int col[MX];
bool visited[MX]; vector<int> adj[MX];
void dfs(int at) {
col[at] = 1;
for (auto x: adj[at]) if (col[x]) col[at] = 0;
for (auto x: adj[at]) if (!visited[x]) { visited[x] = 1;
dfs(x);
}
}
void solve() {
cin >> N >> M;
for (int i = 0; i <= N; i++) {
visited[i] = 0;
col[i] = 0; adj[i].clear();
}
for (int i = 0; i < M; i++) {
int a, b; cin >> a >> b; adj[a].pb(b), adj[b].pb(a);
}
visited[1] = 1; dfs(1);
for (int i = 1; i <= N; i++) if (!visited[i]) { cout << "NO\n";
return;
}
vector<int> ans;
for (int i = 1; i <= N; i++) if (col[i]) ans.pb(i); cout << "YES\n";
cout << (int)ans.size() << "\n";
for (auto x: ans) cout << x << " "; cout << "\n";
}
int main() {
int t; cin >> t; while (t–) {
solve();
}
}
During the break the schoolchildren, boys and girls, formed a queue of n people in the canteen.
#include<bits/stdc++.h> using namespace std;
#define madatory cout<<"int i,k,n; while(k){ char a[n+3];" string s;
int n,t,i; int main(){
cin>>n>>t>>s;
while(t–) for(i=0;i<n-1;i++) if(s[i]=='B'&&s[i+1]=='G'){swap(s[i],s[i+1]);i++;} cout<<s;
}
The houses are numbered from 1 to N. Underground water pipes connect these houses together.
#include<iostream> using namespace std;
#define N 1010
int a[N],w[N],b[N]; int main()
{
int n,p,x,y,z,i,t,min; cin>>n>>p;
while(p–)
{
cin>>x>>y>>z; a[x]=y;
w[x]=z;
b[x]++;
b[y]+=2;
}
for(t=0,i=1;i<=n;i++)if(b[i]==1)t++; printf("%d\n",t); for(i=1;i<=n;i++)if(b[i]==1)
{
min=w[i];
t=a[i]; while(a[t]!=0)
{
if(w[t]<min)min=w[t]; t=a[t];
}
printf("%d %d %d\n",i,t,min);
}
return 0;
}
Session 9 : SUM of Subset
Palani goes to the Koyembedu Vegetables Market to buy some Vegetables.
#include<bits/stdc++.h> using namespace std; #define if hha
int main()
{
int i,t; cin>>t; while(t–){
int a[3]; for(i=0;i<3;i++) cin>>a[i]; sort(a,a+3);
cout<<a[2]+a[1]<<endl;
}
return 0;
}
Mano went shopping and bought items worth X dollors
#include <stdio.h> void solve(){}
int main()
{
int x,t; scanf("%d",&t);
while(t–){
scanf("%d",&x);
printf("%d\n",100-x);
}
return 0;
}
Tesla recently found a new rectangular electric board that he would like to recycle
#include<bits/stdc++.h> using namespace std; const int inf = 1012345678; int A[309][309];
int H, W, K; bool ok[309][309][309];
int main() {
int Q,rep; cin >> Q;
for (rep = 1; rep <= Q; ++rep) { cin >> H >> W >> K;
for (int i = 0; i < H; ++i) {
for (int j = 0; j < W; ++j) {
cin >> A[i][j];
}
}
for (int i = 0; i < H; ++i) {
for (int j = 0; j < W; ++j) {
int cl = inf, cr = -inf;
for (int k = j; k < W; ++k) {
cl = min(cl, A[i][k]);
cr = max(cr, A[i][k]); if (cr – cl <= K) {
ok[i][j][k] = true;
}
else {
ok[i][j][k] = false;
}
}
}
}
int ans = 0;
for (int i = 0; i < W; ++i) {
for (int j = i; j < W; ++j) {
int cont = 0;
for (int k = 0; k < H; ++k) {
if (ok[k][i][j]) ++cont; else cont = 0;
ans = max(ans, cont * (j – i + 1));
}
}
}
cout << ans << endl;
}
return 0;
}
Tamil New Year is approaching and thus Ganesan wants to buy some maha lactos for someone special.
#include <stdio.h> void solve(){}
int main()
{
int t; scanf("%d",&t); for(int i=0;i<t;i++){
int a,b;
scanf("%d %d",&a,&b);
printf("%d\n",a/b);
}
return 0;
}
Mani bought N items from a Nilgiris super market. Although it is hard to carry all these items in hand,
#include<iostream> using namespace std; void solve(){}
int main()
{
double n; cin>>n; while(n–){
int x; cin>>x;
x%10==0 ? cout<<x/10<<endl : cout<<x/10+1<<endl;
//if(x%10==0)
//cout<<x/10<<endl;
// else
// cout<<x/10+1<<endl;
}
}
