Codeforces Round #587 (Div. 3)
Codeforces Round #587 (Div. 3)
A. Prefixes
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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 n, cnt;
string s;
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 >> s;
for (int i = 0; i < n; i += 2){
if (s[i] == s[i + 1]){
cnt ++ ;
if (s[i] == 'b')
s[i] = 'a';
else
s[i] = 'b';
}
}
cout << cnt << "\n" << s << "\n";
return 0;
}
B. Shooting
-
思路：贪心 写个struct 从大到小排序计算
-
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;
struct node{
int a, id;
}a[N];
inline bool cmp(const node &a, const node &b){
return a.a > b.a;
}
int n, 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 >> n;
for (int i = 1; i <= n; i ++ ){
cin >> a[i].a;
a[i].id = i;
}
sort(a + 1, a + n + 1, cmp);
for (int i = 1; i <= n; i ++ )
ans += a[i].a * (i - 1) + 1;
cout << ans << "\n";
for (int i = 1; i < n; i ++ )
cout << a[i].id << " ";
cout << a[n].id << "\n";
return 0;
}
C. White Sheet
-
思路：把所有不可能的情况枚举出来
-
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 x1, y1_, x2, y2, x3, y3, x4, y4, x5, y5, x6, y6;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> x1 >> y1_ >> x2 >> y2 >> x3 >> y3 >> x4 >> y4 >> x5 >> y5 >> x6 >> y6;
if ((x1 >= x3 && x2 <= x4 && y1_ >= y3 && y2 <= y4) || (x1 >= x5 && x2 <= x6 && y1_ >= y5 && y2 <= y6))
cout << "NO\n";
else if ((y1_ >= y3 && y2 <= y4 && y1_ >= y5 && y2 <= y6) && (x1 >= x3 && x2 <= x6 && x4 >= x5 || x1 >= x5 && x2 <= x4 && x3 <= x6))
cout << "NO\n";
else if ((x1 >= x3 && x2 <= x4 && x1 >= x5 && x2 <= x6) && (y1_ >= y3 && y2 <= y6 && y4 >= y5 || y1_ >= y5 && y2 <= y4 && y3 <= y6))
cout << "NO\n";
else
cout << "YES\n";
return 0;
}
D. Swords
-
思路：排序后求最大的和其他每一项的差之和与gcd 答案为\(sum / gcd\)和\(gcd\)
-
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 = 2e5 + 10;
int n;
ll sum, gcd_;
ll a[N];
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; i ++ )
cin >> a[i];
sort(a + 1, a + n + 1);
for (int i = 1; i < n; i ++ ){
sum += a[n] - a[i];
gcd_ = gcd(gcd_, a[n] - a[i]);
}
cout << sum / gcd_ << " " << gcd_ << "\n";
return 0;
}
E1. Numerical Sequence (easy version)
-
思路：只会easy version 暴力搞
-
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 q, tot;
ll k;
string s;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> q;
while (q -- ){
tot = 1;
s.clear();
cin >> k;
while (s.length() < k){
k -= s.length();
s += to_string(tot ++ );
}
cout << s[k - 1] << "\n";
}
return 0;
}
