2020 CDUT寒假集训第一场（思维+dp专场）
2020 CDUT寒假集训第一场（思维+dp专场）
A - Vasya and Golden Ticket
-
思路：暴力
-
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, sum, div_, cnt, tmp;
bool flag;
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 (auto a: s)
sum += a - '0';
for (int i = 2; i <= n; i ++ ){
cnt = 0, tmp = 0;
div_ = sum / i;
if (sum % i == 0){
for (int j = 0; j < n; j ++ ){
if (tmp < div_)
tmp += (s[j] - '0');
if (tmp == div_){
tmp = 0;
cnt ++ ;
}
if (cnt == i && j == n - 1){
cout << "YES\n";
return 0;
}
if (tmp > div_)
break;
}
}
}
cout << "NO\n";
return 0;
}
B - Slime
-
思路：思维题 \(ans\)为排序后的\((a_n-a_1)+\sum_{i=2}^n abs(a_i)\)
-
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 = 5e5 + 10;
const ll INF = 1e18;
int n;
ll ans;
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;
if (n == 1){
ll x;
cin >> x;
cout << x << "\n";
return 0;
}
for (int i = 1; i <= n; i ++ )
cin >> a[i];
sort(a + 1, a + n + 1);
ans = a[n] - a[1];
for (int i = 2; i < n; i ++ )
ans += abs(a[i]);
cout << ans << "\n";
return 0;
}
E - Multiplicity
-
思路：枚举\(a_i\)的因数 可以将二维dp压成一维 \(dp_i = dp_i + dp_{i-1}\)
-
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;
const int mod = 1e9 + 7;
int n, ans;
int a[N], dp[N];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
dp[0] = 1;
cin >> n;
for (int i = 1; i <= n; i ++ ){
cin >> a[i];
for (int j = sqrt(a[i]); j >= 1; j -- ){
if (a[i] % j == 0){
int div = a[i] / j;
if (div <= n)
dp[div] = ((dp[div] + dp[div - 1]) % mod + mod) % mod;
if (j <= n && j != div)
dp[j] = ((dp[j] + dp[j - 1]) % mod + mod) % mod;
}
}
// for (int j = 1; j <= n; j ++ )
// cout << dp[j] << "\n";
}
for (int i = 1; i <= n; i ++ )
ans = ((ans + dp[i]) % mod + mod) % mod;
cout << ans << "\n";
return 0;
}
G - Make Product Equal One
-
思路：水题 没开ll wa1
-
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;
}
ll n, a, ans, num_0, num_neg;
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;
if (a == 0)
num_0 ++ ;
else if (a > 0)
ans += a - 1;
else{
ans += -1 - a;
num_neg ++ ;
}
}
if (num_neg % 2 && num_0 == 0)
ans += 2;
cout << ans + num_0 << "\n";
return 0;
}
H - Restricted RPS
-
思路：模拟
-
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, n, a, b, c, 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 >> t;
while (t -- ){
string ans;
cnt = 0;
cin >> n >> a >> b >> c >> s;
for (int i = 0; i < n; i ++ ){
ans += '#';
if (s[i] == 'R' && b){
ans[i] = 'P';
b -- ;
cnt ++ ;
}
else if (s[i] == 'P' && c){
ans[i] = 'S';
c -- ;
cnt ++ ;
}
else if (s[i] == 'S' && a){
ans[i] = 'R';
a -- ;
cnt ++ ;
}
}
if (cnt < (n + 1) / 2)
cout << "NO\n";
else{
cout << "YES\n";
for (int i = 0; i < n; i ++ ){
if (ans[i] == '#'){
if (a){
ans[i] = 'R';
a -- ;
}
else if (b){
ans[i] = 'P';
b -- ;
}
else if (c){
ans[i] = 'S';
c -- ;
}
}
}
cout << ans << "\n";
}
}
return 0;
}
