prompt string | response string |
|---|---|
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include<iostream>
#include<algorithm>
using namespace std;
int main(){
int n,m;
while(cin>>n>>m,n){
int veg[n];
for(int i=0;i<n;i++)cin>>veg[i];
sort(veg,veg+n,greater<int>());
int res=0;
for(int i=0;i<n;i++){
if((i+1)%m==0)continue;
res+=veg[i];
}
cout << res << endl;
}
... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include<iostream>
#include<algorithm>
using namespace std;
int n, m, x[1000000];
long long sum1, sum2;
int main() {
while (true) {
sum1 = 0; sum2 = 0;
cin >> n >> m;
if (n == 0 && m == 0) { break; }
for (int i = 0; i < n; i++) {
cin >> x[i];
sum1 += x[i];
}
sort(x, x + n);
for (int i = n - m; i >=... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
int main(){
int n, m;
while(cin >> n >> m, n || m){
int price[10010];
int ans = 0;
for(int i=1; i <= n; i++)
cin >> price[i];
sort(price+1, price+n+1, greater<int>());
for(int i=1; i <= n; i++){
if(i % m == 0) continue... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <iostream>
#include <algorithm>
using namespace std;
int main(){
int n,m;
while(cin >> n >> m , n + m){
int yasai[1000] = {0};
for(int i = 1 ; i <= n ; i++){
cin >> yasai[i];
}
sort(yasai,yasai+(n+1),greater<int>());
for(int i = n ; i >0 ; i--){
yasai[i] = yasai[i-1];
... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include<iostream>
#include<queue>
#define loop(i,a,b) for(int i=a;i<b;i++)
#define rep(i,a) loop(i,0,a)
using namespace std;
int main(){
int n,m,dt;
while(1){
cin>>n>>m;
if(!(n&&m))break;
priority_queue<int> que;
rep(i,n){cin>>dt;que.push(dt);}
int cnt=0,sum=0;
while(!que.empty()){
cnt++;
if(cnt!=... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
try (Scanner in = new Scanner(System.in)) {
solve(in);
}
}
private static void solve(Scanner in) {
// n m
// p1 p2 ... pn
while (true) {
... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int n, m;
int veg[1024];
int sum;
while (1){
sum = 0;
scanf("%d%d", &n, &m);
if (n == 0 && m == 0){
break;
}
for (int i = 0; i < n; i++){
scanf("%d", &veg[i]);
}
sort(veg, veg + n);
reverse(veg, ... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <iostream>
#include <algorithm>
using namespace std;
int main() {
int n, m;
int data[1000];
int sum;
while(1) {
cin >> n >> m;
if(n == 0 && m == 0) break;
for(int i = 0; i < n; i++) {
cin >> data[i];
}
sort(data, data+n);
sum = 0;
int i = n-1;
while(1) {
f... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <bits/stdc++.h>
using namespace std;
#define for_(i,a,b) for(int i=(a);i<(b);++i)
int n, m, p[1010];
int main() {
while (cin >> n >> m, n) {
for_(i,0,n) cin >> p[i];
sort(p, p + n, greater< int >());
int ans = 0, cnt = 1;
for_(i,0,n) {
if (cnt == m) {
cnt = 1;
} else {
++cnt;... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include<stdio.h>
int main(void)
{
int a,s,d[2000],f,g,h,i,j;
scanf("%d %d",&a,&s);
while(a!=0&&s!=0){
for(i=1;i<=a;i++){
scanf("%d",&d[i]);
}
for(i=1;i<=a;i++){
for(j=i;j<=a;j++){
if(d[i]<d[j]){
f=d[i];
d[i]=d[j];
d[j]=f;
}
}
}
g=0;
h=0;
for(i=1;i<=a;i++){
if(g==s-1)... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | while True:
n, m = map(int, input().split())
if n == 0:
break
plst = sorted(list(map(int, input().split())), reverse=True)
s = sum(plst)
for i in range(m - 1, n, m):
s -= plst[i]
print(s)
|
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | # AOJ 0227: Thanksgiving
# Python3 2018.6.24 bal4u
while 1:
n, m = map(int, input().split())
if n == 0: break
p = list(map(int, input().split()))
p.sort(reverse=True)
ans = 0
for i in range(1, n+1):
if i % m: ans += p[i-1]
print(ans)
|
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int main(void){
int n,m;
while(cin>>n>>m, n||m){
vector<int> p(n);
for(int i=0;i<n;++i) cin>>p[i];
sort(p.begin(), p.end(), greater<int>());
int sum=0,tmp=0;
for(int i=0;i<n-(n%m);++i){
tmp+=p[i];
if(i%m=... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int main(void)
{
while (1)
{
int n, m, p[1000], sum = 0;
cin >> n >> m;
if (n == 0 && m == 0)break;
for (int i = 0; i < n; i++)
{
cin >> p[i];
sum += p[i];
}
sort(p, p + n);
for (int i = n - m; i >= 0; i -= m)
{
s... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int N,M;
while(cin >> N >> M,(N||M))
{
int a[1000],ans=0;
for(int i=0; i<N; i++)
{
cin >> a[i];
ans+=a[i];
}
sort(a,a+N);
reverse(a,a+N);
for(int i=M-1; i<N; i+=M)
ans-=a[i];
cout << ans << endl;
}
} |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <iostream>
#include <queue>
#include <algorithm>
using namespace std;
int main(){
int m,n,a,b,sum, temp;
int ans;
priority_queue<int> que;
while(1) {
sum = 0;
cin>>n>>m;
if(n == 0 && m == 0){
break;
}
for(int i = 0; i<n; ++i){
cin>>a;
que.push(a);
}
f... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | import java.util.Arrays;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n, m;
while ((n = in.nextInt()) != 0 && (m = in.nextInt()) != 0) {
int[] yasai = new int[n];
for (int i = 0; i < n; i++) {
yasai[i] = in.nextInt();
... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int main(){
int m,n;
int b[1001];
while(cin >> n >> m,m || n){
int total=0;
int k=n/m;
for(int i=0;i<n;i++) cin >> b[i];
sort(b,b+n);
for(int i=1;i<=n;i++){
if(i%m) total+=b[n-i];
}
cout << total << end... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | #include <iostream>
#include <vector>
#include <algorithm>
int main(){
int n,m; //n:buy count m:pack count
while(std::cin >> n >> m){
if(n == 0 && m == 0)
break;
std::vector<int> price;
price.resize(n);
for(int i=0; i<n; ++i){
std::cin >> price[i];
}
std::sort(price.begin(), price.end(), s... |
Problem: As bad weather continues and vegetable prices soar, Seven-Eleven is offering customers bulk purchase sales of vegetables. The store is very busy, as you can get vegetables that are hard to find in stores at reasonable prices.
One day, a group of three good friends living in the Matsunaga housing complex bloom... | while True:
try:
__, m = (lambda row: [int(row[i]) for i in range(len(row))])(input().split())
price = sorted(list(map(int, input().split())), reverse=True)
print(str(sum([price[i] for i in range(len(price)) if (i+1)%m != 0])))
except:
break
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
#include<vector>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<stack>
#include<queue>
#include<math.h>
using namespace std;
typedef long long ll;
#define int long long
typedef vector<int> VI;
typedef pair<int, int> pii;
#define fore(i,a) for(auto &i:a)
#define REP(i,n) fo... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
int res = 0;
int row = 1;
int w = 0;
while ( n >= row ) {
// cout << w << " " << row << endl;
if ( row*k >= w ) {
++res;
w += row;
n -= row;
}
else if ( n >= row+1 ) ++row;
else break;
}
... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | n,k = map(int, input().split())
res = 0
row = 1
w = 0
while n >= row :
if row*k >= w :
res += 1
w += row
n -= row
elif n >= row+1 : row += 1
else : break
print(res)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | import java.io.*;
import java.util.*;
class Main {
void solve(){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt(), k = sc.nextInt();
int res = 0;
int row = 1;
int w = 0;
while ( n >= row ) {
if ( row*k >= w ) {
++res;
w += row;
n -= row;
}
else if ( n >= row+1 ) ++row;
e... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
int N, K;
int main() {
int h, m;
double rest;
h = 1;
cin >> N >> K;
rest = N;
while (true) {
m = rest / (K + 1) * K;
if (m == 0) {
break;
}
++h;
rest = m;
}
cout << h << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include "bits/stdc++.h"
#define rep(var,cnt) for(int (var)=0; (var)<(cnt); ++(var))
using namespace std;
using int64 = int64_t;
using pi=pair<int,int>;
template <int mod>
struct ModInt {
int x;
ModInt() : x(0) {}
ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
ModInt &operator+=(const Mod... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | N,K = map(int, input().split())
t = n = a = 1
while True:
n = (K+a-1)//K
a += n;
if N < a: break
t += 1
print(t)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
using namespace std;
int main(){
int N, K;
cin >> N >> K;
int n = N;
int sum = 0;
int ans = 0;
while (n >= 0) {
int row = (sum + K - 1) / K;
row = max(row, 1);
// cout << n << " " << sum << " " << ans << " " << row << endl;
if (row > n)
... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
#include <algorithm>
using namespace std;
class Solver {
public:
static void solve() {
int n, k;
cin >> n >> k;
int now = 1;
n--;
int sum = 1;
while (true) {
int need = sum / k;
if (sum % k > 0)need++;
if (n < need)break;
now++;
sum += need;
n -= need;
}
cout ... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | import java.util.*;
public class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int count = sc.nextInt();
int strength = sc.nextInt();
int total = 1;
int size = 1;
count--;
while (true) {
int x = (total + strength - 1) / strength;
if (count >= x) {
... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
using namespace std;
int main(void)
{
int N, K;
cin >> N >> K;
int cnt = 0;
while(N>0){
int x = (N-1)/(K+1)+1;
N -= x;
cnt++;
}
cout << cnt << endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | n, k = map(int, input().split())
ans = 1
w = 1
while n-w>0:
blw = (w+k-1)//k
w += blw
if n-w>=0: ans+=1
print(ans)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
using namespace std;
int main(void){
int n, k;
cin >> n >> k;
int x = 1, y = 0, sum = 0;
while(1){
if(sum + x > n) break;
if(sum > k * x){
x = sum / k;
if(sum % k > 0) x++;
}
else sum += x, y++;
}
cout << y << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int INF = 1<<30;
const int MOD = 1e9 + 7;
cons... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
#define INCANT cin.tie(0), cout.tie(0), ios::sync_with_stdio(false), cout << fixed << setprecision(20);
#define int long long
#define gcd __gcd
#define all(x) (x).begin(), (x).end()
const int INF = 1e18, MOD = 1e9 + 7;
template<class T>
bool chmax(T& a, T b){return (a = max... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
int main(){
int n,k,sum=0,now=1,ans=0;
cin>>n>>k;
for(;;){
if(sum+now>n)break;
else if(sum<=now*k)sum+=now,ans++;
else now++;
}
cout<<ans<<endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ll n, k; cin >> n >> k;
int ans = 0, sum = 0;
while(n >= max(1LL, (sum + k - 1) / k)) {
ans += 1;
const ll add = max(1LL, (sum + k - 1) / k);
sum += add;
n -= add;
}
cout << ans << en... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
using namespace std;
#define rep(i,x,y) for(int i=x;i<y;i++)
#define print(a,n) rep(i,0,n){ cout<<(i ? " ":"")<<a[i]; }cout<<endl;
#define pprint(a,m,n) rep(j,0,m){ print(a[j],n); }
const int mod = 1e9+7;
const int size=1e5;
const int INF=1e9;
int main(){
int N,K;cin>>N>>K;
int sum=1;int i=1;
... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
int judge(int n,int k,int amari){
if(amari<=1){
return 0;
} else {
for(double i=2;;i++){
if((n-amari)/i<=k) return i;
if(amari<=i) return 0;
}
}
}
int main(){
int n,k;
cin >> n >> k;
int num=k+1;
in... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
using namespace std;
int putBlocks(int N, int K)
{
if (N == 1) return 1;
else {
int most_under = 1;//最下段のブロックの数
int count = N;//注目塔の高さ
while ((((double)count - 1) / (double)most_under) > K) {//注目塔の最下段の強度を判定
most_under++;
count--;//塔の高さを減らす（最下段へ）
}
return 1 + putBlocks(cou... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include "bits/stdc++.h"
using namespace std;
int main() {
long long N, K, NOW = 1, ANS = 0;
cin >> N >> K;
while (N >= NOW) {
NOW += (NOW + K - 1) / K;
ANS++;
}
cout << ANS << endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}
//INSERT ABOVE HERE
signed main(){
Int n,k;
cin>>n>>k;
Int ans=1,sum=1;
n--;
while(1)... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using LL = long long;
using namespace std;
int main(){
int N,K;cin >> N >> K;
LL ans = 1;
LL weight = 1;
LL ins = 0;
N--;
while(N != 0){
// cout<<weight<<endl;
if(ins == 0){
ins++;
N--;
}else
if(1.0*weight/ins <= K)... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #define _AOJ_
#include "bits/stdc++.h"
using namespace std;
typedef long long i64;
#define rep(i,n) for(int i=0;i<n;++i)
#define REP(i,a,b) for(int i=a;i<b;++i)
#define all(c) c.begin(),c.end()
//define pfs,bellmanford,etc.
int main(){
i64 n,k;cin>>n>>k;
i64 s=1,ans=1;
vector<i64>a(1,1);
while(s<=n){
a.push_bac... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
int main() {
int n,k,t=0,ans=0,u=1;
cin >> n >> k;
while (true) {
if (n-t<u) break;
if (u*k>=t) {t+=u;ans++;}
else {u++;continue;}
}
cout << ans << endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | import math
n, k = map(int, input().split())
ret = 1
acc = 1
while True:
want = math.ceil(acc / k)
if(want+acc > n):
break
acc += want
ret += 1
print(ret)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> mp;
#define inf 1e9
int main(){
int n,k;
cin>>n>>k;
int sum = 0;
int h = 0;
int y = 1;
while(n-y>=0){
// cout<<n<<endl;
if((double)sum/y <= k){
sum+=y;
n-=y;
h++;
}else{
while( ( (double) sum/y) > (double... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;
int main(){
int n,K;cin>>n>>K;
int cnt=1,i=1;
for(;;i++){
int need=(cnt+K-1)/K;
if(cnt+need>n)break;
cnt+=need;
}
cout<<i<<endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<int, ll> pil;
typedef pair<int, ll> pli;
typedef pair<double,double> pdd;
#define SQ(i) ((i)*(i))
#define MEM(a, b) memset(a, (b), sizeof(a))
#define SZ(i) int(i.size())
#define FOR(i, ... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<iostream>
using namespace std;
int main(){
int N, K, count = 0, sum = 0, num = 1;
cin>>N>>K;
for(;;){
if(sum > num*K){
num = (sum+K-1)/K;
//cout<<"num = "<<num<<endl;
}
N -= num;
sum += num;
if(N < 0)break;
count++;
//cout<<count<<" "<<sum<<" "<<num<<endl;
}
cout<<count<<endl;
... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
using namespace std;
int main(){
int N ,K;
cin >> N >> K;
int ans = 1;
int wei = 1;
N--;
while(N > 0){
int n = ceil((double)wei / (double)K);
if(N >= n) ans++;
wei += n;
N -= n;
}
cout << ans << endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
using llong = long long;
using ldbl = long double;
using P = pair<llong, llong>;
#define BE(x) x.begin(), x.end()
const llong inf = llong(1e18)+7;
const llong mod = 1e9+7;
int main(){
llong N, K;
cin >> N >> K;
llong now = 1, ans = 1;
N--;
while(true){
llong next... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<iostream>
#include<cassert>
using namespace std;
int solve(int N, int K){
int t = 1;
int n = 1;
int a = 1;
while(1){
n = (K+a-1)/K;
a += n;
if ( N < a ) return t;
t++;
}
}
int main() {
int N, K; cin >> N >> K;
int t = solve(N, K);
cout << t << endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
#include <assert.h>
using namespace std;
int main()
{
int n, k;
cin >> n >> k;
n--;
int sum = 1, ans = 1;
while ((sum + k - 1) / k <= n)
{
n -= (sum + k - 1) / k;
sum += (sum + k - 1) / k;
ans++;
}
cout << ans << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define HUGE_NUM 99999999999999999
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;
int main(){
int N,K;
scanf("%d %d",&N,&K);
int ans = 1,upper = 1;
int rest = N-1,need;
while(t... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k; cin >> n >> k;
int remove = 0, count = 0;
if( n <= k + 1 ) {
cout << n << endl;
} else {
n -= k + 1;
remove += k + 1;
while( 1 ) {
n -= ceil( static_cast<double>( remove ) / k );
remove += ceil( static_cast<double>( remove ) / k );... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <cstdio>
#include <vector>
#include <algorithm>
#include <utility>
bool valid(int N, int K, int x) {
int wsum = 1;
for (int i = 1; i < x; ++i) {
int n = (wsum+K-1) / K;
wsum += n;
if (wsum > N) return false;
}
return wsum <= N;
}
int main() {
int N, K;
scanf("%d %d", &N, &K);
int l... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<iostream>
using namespace std;
int main(){
int n, k; cin >> n >> k;
k++;
int ans = 0;
while (n > 0) {
int m = (n + k - 1) / k;
n -= m;
ans++;
}
cout << ans << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
long base = sc.nextLong(), K = sc.nextLong();
long next;
int ans = 1;
while (base >= 1) {
long low = 0, high = base;
while (hig... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
using namespace std;
int main(){
int n,k,x=0,sum=0,a=1;
cin>>n>>k;
while(1){
if(sum>=n) break;
sum+=a;
if((sum-a)/a>=k){
sum-=a; ++a; --x;
}
++x;
}
cout<<x<<endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using VI = vector<int>;
using VL = vector<ll>;
#define FOR(i,a,n) for(int (i)=(a);(i)<(n);(i)++)
#define eFOR(i,a,n) for(int (i)=(a);(i)<=(n);(i)++)
#define rFOR(i,a,n) for(int (i)=(n)-1;(i)>=(a);(i)--)
#define erFOR(i,a,n) for(int (i)=(n);(i)>=(a);(i)... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | // 2018/12/09 Tazoe
#include <iostream>
using namespace std;
int main()
{
int N, K;
cin >> N >> K;
int cnt = 0;
while(N>0){
int x = (N-1)/(K+1)+1;
N -= x;
cnt++;
}
cout << cnt << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
int main(){
int n, k;
cin >> n >> k;
int ans = 1, t = 1;
while((t += (t+k-1)/k) <= n) ans++;
cout << ans << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
int ret = 1;
int used = 1;
while (true) {
used += (used + k - 1) / k;
if (used > n)break;
ret++;
}
cout << ret << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | N, K = map(int, input().split())
ans = 1
s = 1
N -= 1
while 1:
m = (s+K-1) // K
if N < m:
break
s += m; N -= m
ans += 1
print(ans)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | N, K = (int(x) for x in input().split())
t = 1
r = N - 1
s = 1
while True:
b = s // K + 1 if s % K else s // K
if r >= b:
r -= b
s += b
t += 1
else:
break
print(t)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
int main()
{
int N, K;
int cnt = 0;
int num = 0;
cin >> N >> K;
for ( int i = 0; i < N; i++ ) {
num++;
if ( num * K >= N - i - 1 ) {
num = 0;
cnt++;
}
}
cout << cnt << endl;
return ( 0 );
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... |
N,K = [int(i) for i in input().split()]
weight = 1
ans = 1
N -= 1
while N > 0:
d = 0
if weight % K == 0:
d = weight // K
else:
d = weight // K + 1
N -= d
weight += d
if N >= 0:
ans += 1
print(ans)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#define int long long
using namespace std;
signed main() {
int n, k; cin >> n >> k;
int sum = 0, ans = 0;
while (sum < n) {
int p = (sum + k - 1) / k;
if (!sum)p = 1;
if (sum + p > n)break;
sum += p;
ans++;
}
cout << ans << endl;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #define _USE_MATH_DEFINES
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
using namespace std;
typedef pair<long long int, long long int> P;
long long int INF = 1e18;
long long int MOD = 1e9 + 7;
int... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | //yukicoder@cpp14
//author:luckYrat(twitter:@luckYrat_)
//<ここに一言>
//せんげん！
#include <iostream>
#include <cmath>
#include <algorithm>
#include <iomanip>
#include <string>
#include <vector>
#include <set>
#include <stack>
#include <queue>
#include <map>
#include <bitset>
#include <cctype>
#include <utility>
#include <... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
#define mod 1000000007
int main(){
ll n,k; cin>>n>>k;
ll ans=0,sum=0;
while(1){
ll need=(sum+k-1)/k;
if(need==0)need=1;
if(n<need)break;
n-=need;
sum+=need;
ans++;
}
cout<<an... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<cstdio>
#include<functional>
#include<algorithm>
using namespace std;
int main(void)
{
int n,k,x,sum,cnt;
scanf("%d %d",&n,&k);
cnt=1; sum=1; n--;
while(1) {
x=sum/k+(sum%k!=0);
if(x>n) break;
cnt++; n-=x; sum+=x;
}
printf("%d\n",cnt);
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | n,k = map(int,input().split())
ans = 0
d,s = 1,0
while True:
# s/d <= k
# s <= d*k
while d*k < s:
d += 1
if s+d > n:
break
s += d
ans += 1
print(ans)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
using namespace std;
int main()
{
int n, k, m = 0, i = 1, j = 0;
cin >> n >> k;
while (1) {
while (j > k * i)
i++;
if (n - j < i)
break;
m++;
j += i;
}
cout << m << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<iostream>
#include<vector>
#include<set>
#include<algorithm>
#include<queue>
#include<utility>
#define LL long long
using namespace std;
int main() {
int n, _n, k;
cin >> n >> k;
int sum = 1, ans = 1;
_n = n - 1;
while ((sum + k - 1) / k <= _n) {
_n -= (sum + k - 1) / k;
sum += (sum + k - 1) / k;
... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int main() {
cin.tie(0);
ios::sync_with_stdio(false);
ll N,K; cin >> N >> K;
ll ans = 1;
ll sum = 1;
N--;
while(true){
if(N*K < sum) break;
ll tmp = 0;
if(sum%K == 0) tmp... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k, m=1, sum=0, ans=0;
cin >> n >> k;
while(n>=m){
ans++;
sum += m;
n -= m;
m = (sum-1)/k+1;
}
cout << ans << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | n, k = map(int, input().split())
weight = 1
rest = n - 1
layers = 1
while True:
add = weight // k + bool(weight % k)
if add <= rest:
rest -= add
weight += add
layers += 1
else:
break
print(layers)
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <iostream>
#include <complex>
#include <string>
#include <algorithm>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <functi... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
#include <vector>
using std::vector;
using std::cin;
using std::cout;
using std::endl;
int main()
{
int N, K, num = 0; // N: ブロック個数、K: ブロック強度、num: 積み上げるブロック数
vector<int> digit;
cin >> N >> K;
for (;;) {
int n = 0;
while (num > ++n * K);
num += n;
if (num > N) {
break;
}
digit.p... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include<bits/stdc++.h>
#define V vector
#define VI vector<int>
#define VVI vector<vector<int>>
#define rep(i,n) for(int i=0;i<(n);i++)
#define MOD 1000000007
using namespace std;
int main(void){
int N,K;
cin>>N>>K;
int t=1,n=1,a=1;
while(1){
n=(K+a-1)/K;
a+=n;
if(N<a){
... |
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
n--;
int result = 1;
int used = 1;
int use;
while (n >= 0) {
result++;
use = (used + k - 1) / k;
used += use;
n -= use;
}
cout << result - 1 << endl;
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include <iostream>
using namespace std;
long long N, K;
int main() {
cin >> N >> K;
long long ret = 1, cnt = 1;
while (true) {
long long nex = (ret + K - 1) / K;
if (ret + nex > N) {
cout << cnt << endl;
return 0;
}
ret += nex;
cnt++;
}
return 0;
}
|
Problem: We make a tower by stacking up blocks. The tower consists of several stages and each stage is constructed by connecting blocks horizontally. Each block is of the same weight and is tough enough to withstand the weight equivalent to up to $K$ blocks without crushing.
We have to build the tower abiding by the f... | #include "bits/stdc++.h"
#pragma warning(disable:4996)
using namespace std;
int main() {
int N,K;cin>>N>>K;
int ans=0;
int sum=0;
while (true) {
int next=(sum-1)/K+1;
if(ans==0)next=1;
if (N < sum + next) {
break;
}
else {
sum+=next;
ans++;
}
}
cout<<ans<<endl;
return 0;
}
|
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include <iostream>
#include <vector>
using namespace std;
int main(){
int n,k,t;
bool f=true;
vector<int> s;
while(1){
cin >> n >> k;
if(n==0&&k==0) break;
for(int i=0;i<k;i++){
cin >> t;
s.push_back(t);
}
for(int i=0;i<n;i++){
for(int j=0;j<k;j++){
cin >> t;
s[j]-=t;
}
}
fo... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include <iostream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <sstream>
#include <map>
#include <set>
#define REP(i,k,n) for(int i=k;i<n;i++)
#define rep(i,n) for(int i=0;i<n;i++)
#define INF 1<<30
#define pb push_back
#define mp make_pair
using namespace std;
typedef long l... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | import java.io.*;
public class Main{
public static void main(String[] args)throws IOException{
BufferedReader r=new BufferedReader(new InputStreamReader(System.in));
while(true){
String s=r.readLine();
String[] t=s.split(" ");
int[] n=new int[t.length];
for(int i=0;i<2;i++){
n[i]=Integer.pa... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include<stdio.h>
int s[100];
int main(){
int n,k;
int i,j;
while(1){
scanf("%d %d",&n,&k);
if(n==0)return 0;
for(i=0;i<k;i++)scanf("%d",&s[i]);
for(i=0;i<n;i++){
for(j=0;j<k;j++){
int x;
scanf("%d",&x);
s[j]-=x;
}
}
for(i=0;i<k;i++)if(s[i]<0)break;
if(i==k)printf("Yes\... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include<iostream>
#include<sstream>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<algorithm>
#include<numeric>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cassert>
#define rep(i,n) for(int i=0;i<n;i++)
#define all(c) (c).begin(),(c).end()
#define mp make_pair
... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n, k;
int s[];
int a[][];
boolean f;
while (true) {
n = sc.nextInt();
k = sc.nextInt();
if ((n | k) == 0) {
break;
}
s = new int[k];
for (int i = 0; i < k... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include<iostream>
#include<string>
using namespace std;
int main(){
while(true){
int N,K;
string ans = "Yes";
cin >> N >> K;
int S[K];
if( N == 0 && K == 0 ) break;
for(int i = 0; i < K; ++i){
cin >> S[i];
}
for(int i = 0; i < N; ++i){
for(int j = 0; j < K; ++j){
int B;... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include <iostream>
using namespace std;
int main() {
int n, k;
while (cin >> n >> k) {
if (n == 0 && k == 0) {
break;
}
int s[100];
bool hantei = true;
for (int i = 0; i < k; i++) {
cin >> s[i];
}
int a;
for (int i = 0; i < n; i++) {
for (int j = 0; j < k; j++) {
cin >> a;
s[j] -= ... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include<iostream>
using namespace std;
int main(){int n,k,l,i;while(1){cin>>n>>k;if(n==0&&k==0)break;int s[k];for(i=0;i<k;i++)s[i]=0,cin>>s[i];for(i=0;i<n;i++)for(int j=0;j<k;j++)cin>>l,s[j]-=l;for(i=0;i<k;i++){if(s[i]<0){cout<<"No\n";break;}if(i==k-1)cout<<"Yes\n";}}} |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include<bits/stdc++.h>
using namespace std;
int N, K;
int M[110];
int main()
{
while(cin >> N >> K && N && K)
{
bool flag = false;
for(int i = 0; i < K; i++)cin >> M[i];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < K; j++)
{
int m;
... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;
using ll=long long;
const int MAX = (int)1e2;
int N, K;
void solve() {
int s[MAX ];
bool ok=0;
int b;
//int b[MAX ][MAX];
for (int i = 0; i < K; i++) {
cin >> s[i];
}
for (int i = 0; i < N; i++) {
for (int j = 0; j < K; j++) {
... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | while True:
N,K=map(int,raw_input().split())
if N==K==0:break
S=map(int,raw_input().split())
L=[0]*K
for i in range(N):
temp=map(int,raw_input().split())
for j in range(K):
L[j]+=temp[j]
for i in range(K):
if S[i]-L[i]<0:
print "No"
bre... |
Problem: There is a vampire family of N members. Vampires are also known as extreme gourmets. Of course vampires' foods are human blood. However, not all kinds of blood is acceptable for them. Vampires drink blood that K blood types of ones are mixed, and each vampire has his/her favorite amount for each blood type.
Y... | #include<iostream>
using namespace std;
int main(){
int N,K,blood_sum[100],blood_num[100],blood;
bool flag;
while(1){
cin >>N>>K;
if(N == 0 && K == 0)break;
flag=0;
for(int i=0 ; i < 100 ; i++)blood_sum[i]=0;
for(int i=0 ; i < K ; i++)cin >>blood_num[i];
for(int i=0 ; i < N ; i++... |
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