Codeforces Round #601 (Div. 2)
Codeforces Round #601 (Div. 2)
A. Changing Volume
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思路：水题
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AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int t;
ll a, b, ans;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
cin >> a >> b;
if (a == b)
cout << "0\n";
else{
ll tmp = abs(a - b);
ans = tmp / 5 + tmp % 5 / 2 + tmp % 5 % 2;
cout << ans << "\n";
}
}
return 0;
}
B. Fridge Lockers
-
思路：贪心
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AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1010;
int t, n, m, ans, res;
struct A{
int a, id;
}a[N];
inline bool cmp(const A &a1, const A &a2){
return a1.a < a2.a;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> t;
while (t -- ){
ans = 0;
cin >> n >> m;
res = m - n;
for (int i = 1; i <= n; i ++ ){
cin >> a[i].a;
a[i].id = i;
}
if (m < n || n == 2){
cout << "-1\n";
continue;
}
sort(a + 1, a + n + 1, cmp);
for (int i = 1; i <= n; i ++ )
ans += a[i].a;
ans *= 2;
for (int i = 1; i <= res; i ++ )
ans += a[1].a + a[2].a;
cout << ans << "\n";
for (int i = 1; i < n; i ++ )
cout << a[i].id << " " << a[i + 1].id << "\n";
cout << a[n].id << " " << a[1].id << "\n";
for (int i = 1; i <= res; i ++ )
cout << a[1].id << " " << a[2].id << "\n";
}
return 0;
}
C. League of Leesins
-
思路：每次更新起点 与起点有关系的点只有两个 然后把与起点有关的点从集合中删除 重新找起点
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1e5 + 10;
int n, tot, q1, q2, q3, beg, en;
int vis[N], ans[N];
bool flag;
set<int> st[N];
void dfs(int x, int cnt){
if (cnt == n - 1)
return ;
for (auto it = st[x].begin(); it != st[x].end(); it ++ )
if (vis[*it] != 2)
st[*it].erase(x);
for (auto it = st[x].begin(); it != st[x].end(); it ++ ){
if (!vis[*it] && st[*it].size() == 2){
ans[ ++ tot ] = *it;
vis[*it] = 1;
dfs(*it, cnt + 1);
}
}
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
for (int i = 1; i <= n - 2; i ++ ){
cin >> q1 >> q2 >> q3;
st[q1].insert(q2), st[q1].insert(q3);
st[q2].insert(q1), st[q2].insert(q3);
st[q3].insert(q1), st[q3].insert(q2);
}
for (int i = 1; i <= n; i ++ ){
if (!flag && st[i].size() == 2){
beg = i;
flag = true;
}
else if (st[i].size() == 2){
en = i;
break;
}
}
ans[ ++ tot ] = beg;
vis[beg] = 1;
for (auto it = st[en].begin(); it != st[en].end(); it ++ )
vis[*it] = 2;
dfs(beg, 1);
for (auto it = st[en].begin(); it != st[en].end(); it ++ )
if (vis[*it] == 2 && st[*it].size() == 4)
ans[ ++ tot ] = *it;
for (auto it = st[en].begin(); it != st[en].end(); it ++ )
if (vis[*it] == 2 && st[*it].size() == 3){
ans[ ++ tot ] = *it;
ans[ ++ tot ] = en;
break;
}
for (int i = 1; i <= tot; i ++ )
cout << ans[i] << " ";
return 0;
}
D. Feeding Chicken
-
思路：类似于蛇形构造 考场降智题意读复杂了
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
#define debug(x) cout << #x << ": " << x << "\n"
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 110;
int t, r, c, k, cnt, tot, x, y, xx;
char mp[N][N];
vector<char> ch;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
for (char i = '0'; i <= '9'; i ++ )
ch.push_back(i);
for (char i = 'A'; i <= 'Z'; i ++ )
ch.push_back(i);
for (char i = 'a'; i <= 'z'; i ++ )
ch.push_back(i);
cin >> t;
while (t -- ){
cin >> r >> c >> k;
cnt = 0, tot = 0;
for (int i = 1; i <= r; i ++ ){
for (int j = 1; j <= c; j ++ ){
cin >> mp[i][j];
if (mp[i][j] == 'R')
cnt ++ ;
}
}
x = cnt / k, y = cnt % k;
xx = x;
if (y){
xx ++ ;
y -- ;
}
for (int i = 1; i <= r; i ++ ){
if (i & 1){
for (int j = 1; j <= c; j ++ ){
if (mp[i][j] == 'R'){
xx -- ;
cnt -- ;
}
mp[i][j] = ch[tot];
if (!cnt){
mp[i][j] = ch[tot];
continue;
}
if (!xx){
if (y){
xx = x + 1;
y -- ;
}
else
xx = x;
tot ++ ;
}
}
}
else{
for (int j = c; j >= 1; j -- ){
if (mp[i][j] == 'R'){
xx -- ;
cnt -- ;
}
mp[i][j] = ch[tot];
if (!cnt){
mp[i][j] = ch[tot];
continue;
}
if (!xx){
if (y){
xx = x + 1;
y -- ;
}
else
xx = x;
tot ++ ;
}
}
}
}
for (int i = 1; i <= r; i ++ ){
for (int j = 1; j <= c; j ++ )
cout << mp[i][j];
cout << "\n";
}
}
return 0;
}
E2. Send Boxes to Alice (Hard Version)
-
思路：参考dalao的思路 枚举sum的质因数x 对于每个x 计算\(a_i\)的前缀和\(sum_i\) 因为要求\(a_i % x = 0\) 所以\(sum_i % x = 0\) 如果\(sum_i % x != 0\) 对余数凑成x 每次取花费最少的然后算总和取最小值
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1e6 + 10;
const ll INF = 1e18;
int n;
int a[N];
ll sum, ans;
void calc(ll x){
ll cnt = 0, now = 0;
for (int i = 1; i <= n; i ++ ){
now = (now + a[i]) % x;
cnt += min(now, x - now);
}
ans = min(ans, cnt);
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
ans = INF;
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> a[i];
sum += a[i];
}
if (sum == 0){
cout << "0\n";
return 0;
}
if (sum == 1){
cout << "-1\n";
return 0;
}
for (ll i = 2; i * i <= sum; i ++ ){
if (sum % i == 0){
calc(i);
while (sum % i == 0)
sum /= i;
}
}
if (sum > 1)
calc(sum);
cout << ans << "\n";
return 0;
}
