Codeforces Round #597 (Div. 2)
A. Good ol' Numbers Coloring
-
思路：水题
-
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, a, b;
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 (gcd(a, b) == 1)
cout << "Finite
";
else
cout << "Infinite
";
}
return 0;
}
B. Restricted RPS
-
思路：照着题意模拟
-
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, cr, cp, cs, cnt, pos;
string s, 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 -- ){
ans.clear();
cr = cp = cs = 0;
cin >> n >> a >> b >> c >> s;
for (int i = 0; i < n; i ++ ){
if (s[i] == 'R')
cr ++ ;
else if (s[i] == 'P')
cp ++ ;
else
cs ++ ;
}
cnt = min(a, cs) + min(b, cr) + min(c, cp);
if (cnt * 2 < n){
cout << "NO
";
continue;
}
for (int i = 0; i < n; i ++ ){
if (s[i] == 'R' && b){
ans += 'P';
b -- ;
}
else if (s[i] == 'P' && c){
ans += 'S';
c -- ;
}
else if (s[i] == 'S' && a){
ans += 'R';
a -- ;
}
else
ans += '#';
}
for (int i = 0; i < n; i ++ ){
if (ans[i] == '#'){
if (a){
ans[i] = 'R';
a -- ;
continue;
}
if (b){
ans[i] = 'P';
b -- ;
continue;
}
if (c){
ans[i] = 'S';
c -- ;
continue;
}
}
}
cout << "YES
" << ans << "
";
}
return 0;
}
C. Constanze's Machine
-
思路：dp 类似于斐波拉契数列
-
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 len;
int dp[N] = {1, 1};
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 >> s;
len = s.length();
for (auto it: s){
if (it == 'w' || it == 'm'){
cout << "0
";
return 0;
}
}
for (int i = 2; i <= len; i ++ ){
dp[i] = dp[i - 1];
if (s[i - 1] == s[i - 2] && (s[i - 1] == 'u' || s[i - 1] == 'n'))
dp[i] = (dp[i - 1] + dp[i - 2]) % mod;
}
cout << dp[len] << "
";
return 0;
}
D. Shichikuji and Power Grid
-
思路：最小生成树
-
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 = 2010;
struct node{
ll x, y, c, k, vis, change;
}num[N];
struct E{
int fir, sec;
}e[N];
int n, tmp, c1, c2;
int p[N];
ll 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 >> num[i].x >> num[i].y;
for (int i = 1; i <= n; i ++ ){
cin >> num[i].c;
num[i].vis = num[i].change = 0;
}
for (int i = 1; i <= n; i ++ )
cin >> num[i].k;
num[0].c = 1 << 30;
for (int i = 1; i <= n; i ++ ){
tmp = 0;
for (int j = 1; j <= n; j ++ )
if (num[j].c < num[tmp].c && !num[j].vis)
tmp = j;
num[tmp].vis = 1;
ans += num[tmp].c;
if (num[tmp].change){
c2 ++ ;
e[c2].fir = tmp;
e[c2].sec = num[tmp].change;
}
else
p[ ++ c1] = tmp;
for (int j = 1; j <= n; j ++ ){
if (num[j].c > (abs(num[j].x - num[tmp].x) + abs(num[j].y - num[tmp].y)) * (num[j].k + num[tmp].k)){
num[j].c = (abs(num[j].x - num[tmp].x) + abs(num[j].y - num[tmp].y)) * (num[j].k + num[tmp].k);
num[j].change = tmp;
}
}
}
cout << ans << "
" << c1 << "
";
for (int i = 1; i <= c1; i ++ )
cout << p[i] << " ";
cout << "
" << c2 << "
";
for (int i = 1; i <= c2; i ++ )
cout << e[i].fir << " " << e[i].sec << "
";
return 0;
}