Codeforces Round #664
A
题意
给四种颜色的球，可以把一个红色一个蓝色一个绿色染成白色，问能不能变成回文串
#include <cstdio>
#include <stack>
#include <set>
#include <cmath>
#include <map>
#include <time.h>
#include <vector>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
//#include <memory.h>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <cassert>
#include <unordered_map>
#define P pair<int, int>
#define LL long long
#define LD long double
#define PLL pair<LL, LL>
#define mset(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i < b; i++)
#define PI acos(-1.0)
#define random(x) rand() % x
#define debug(x) cout << #x << " " << x << "
"
using namespace std;
const int inf = 0x3f3f3f3f;
const LL __64inf = 0x3f3f3f3f3f3f3f3f;
#ifdef DEBUG
const int MAX = 2e3 + 50;
#else
const int MAX = 2e6 + 500;
#endif
const int mod = 1e9 + 7;
void file_read() {
#ifdef DEBUG
freopen("in", "r", stdin);
// freopen("out", "w", stdout);
#endif
}
int main() {
file_read();
int T;
LL a, b, c, d;
scanf("%d", &T);
while (T--)
{
cin >> a >> b >> c >> d;
LL tot = a + b + c + d;
if(tot & 1) {
int odd = (a & 1) + (b & 1) + (c & 1) + (d & 1);
if(odd == 1) {
puts("Yes");
continue;
}
if(a > 0 and b > 0 and c > 0) {
a--, b--, c--, d+=3;
odd = (a & 1) + (b & 1) + (c & 1) + (d & 1);
if(odd == 1) {
puts("Yes");
continue;
}
}
puts("No");
continue;
}
else {
int odd = (a & 1) + (b & 1) + (c & 1) + (d & 1);
if(odd == 0 ) {
puts("Yes");
continue;
}
if(a > 0 and b > 0 and c > 0 and d > 0) {
a--, b--, c--, d += 3;
odd = (a & 1) + (b & 1) + (c & 1) + (d & 1);
if(odd == 0) {
puts("Yes");
continue;
}
}
puts("No");
}
}
return 0;
}
B
题意
在棋盘上走，类似于车，求遍历所有格子一次的方案
#include <cstdio>
#include <stack>
#include <set>
#include <cmath>
#include <map>
#include <time.h>
#include <vector>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
//#include <memory.h>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <cassert>
#include <unordered_map>
#define P pair<int, int>
#define LL long long
#define LD long double
#define PLL pair<LL, LL>
#define mset(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i < b; i++)
#define PI acos(-1.0)
#define random(x) rand() % x
#define debug(x) cout << #x << " " << x << "
"
using namespace std;
const int inf = 0x3f3f3f3f;
const LL __64inf = 0x3f3f3f3f3f3f3f3f;
#ifdef DEBUG
const int MAX = 2e3 + 50;
#else
const int MAX = 2e6 + 500;
#endif
const int mod = 1e9 + 7;
void file_read() {
#ifdef DEBUG
freopen("in", "r", stdin);
// freopen("out", "w", stdout);
#endif
}
void print(int x, int y) {
printf("%d %d
", x, y);
}
bool vis[105][105];
int main() {
file_read();
int n, m, x, y;
int X, Y;
cin >> n >> m >> x >> y;
X = x, Y = y;
print(x, y--);
while(y >= 1) {
print(x, y--);
}
y = Y+1;
while(y <= m) {
print(x, y++);
}
y = m;
for(int i = x-1; i >= 1; i--) {
if(y == m){
for(; y >= 1; y--)
print(i, y);
y = 1;
}
else if(y == 1) {
for(; y <= m; y++)
print(i, y);
y = m;
}
}
for(int i = x+1; i <= n; i++) {
if(y == m){
for(; y >= 1; y--)
print(i, y);
y = 1;
}
else if(y == 1) {
for(; y <= m; y++)
print(i, y);
y = m;
}
}
return 0;
}
C
题意
给两个序列(a),(b), 对每个(a_i)可以选择一个(b_j),(c_i=a_i&b_j),求最小的(c_1|c_2|dots|c_n)
Solution
从高位到低位，使得每一位尽可能为0，维护对于每个(a_i),可以用的(b_j)
#include <cstdio>
#include <stack>
#include <set>
#include <cmath>
#include <map>
#include <time.h>
#include <vector>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
//#include <memory.h>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <cassert>
#include <unordered_map>
#define P pair<int, int>
#define LL long long
#define LD long double
#define PLL pair<LL, LL>
#define mset(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i < b; i++)
#define PI acos(-1.0)
#define random(x) rand() % x
#define debug(x) cout << #x << " " << x << "
"
using namespace std;
const int inf = 0x3f3f3f3f;
const LL __64inf = 0x3f3f3f3f3f3f3f3f;
#ifdef DEBUG
const int MAX = 2e3 + 50;
#else
const int MAX = 2e6 + 500;
#endif
const int mod = 1e9 + 7;
void file_read() {
#ifdef DEBUG
freopen("in", "r", stdin);
// freopen("out", "w", stdout);
#endif
}
int n, m;
LL a[MAX], b[MAX];
bool ok[205][205];
bool check(LL x) {
bool res = true;
for(int i = 1; i <= n; i++) {
bool cur = false;
for(int j = 1; j <= m; j++) {
if(!ok[i][j]) continue;
if((a[i] & b[j] & x) == 0) {
cur = true;
break;
}
}
res &= cur;
}
if(res) {
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= m; j++) {
if((a[i] & b[j] & x) != 0) {
ok[i][j] = false;
}
}
}
}
return res;
}
int main() {
file_read();
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; i++) cin >> a[i];
for(int i = 1; i <= m; i++) cin >> b[i];
LL ans = 0;
mset(ok, 1);
for(LL i = 32; i >= 0; i--) {
if(!check(1LL<<i)) {
ans |= (1LL<<i);
}
}
cout << ans << "
";
return 0;
}
D
题意
给以序列(a),可以选(n)次数字，如果数字大于(m),那么后(d)次就不能选，求选出的数字的最大值
Solution
枚举选的数字小于等于(m)的个数，可以选择超过(m)的数字就是固定的，选择最大的
#include <cstdio>
#include <stack>
#include <set>
#include <cmath>
#include <map>
#include <time.h>
#include <vector>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
//#include <memory.h>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <cassert>
#include <unordered_map>
#define P pair<int, int>
#define LL long long
#define LD long double
#define PLL pair<LL, LL>
#define mset(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i < b; i++)
#define PI acos(-1.0)
#define random(x) rand() % x
#define debug(x) cout << #x << " " << x << "
"
using namespace std;
const int inf = 0x3f3f3f3f;
const LL __64inf = 0x3f3f3f3f3f3f3f3f;
#ifdef DEBUG
const int MAX = 2e3 + 50;
#else
const int MAX = 2e6 + 500;
#endif
const int mod = 1e9 + 7;
void file_read() {
#ifdef DEBUG
freopen("in", "r", stdin);
// freopen("out", "w", stdout);
#endif
}
LL a[MAX];
LL A[MAX], B[MAX];
LL sa[MAX], sb[MAX];
int main() {
file_read();
LL n, d, m;
cin >> n >> d >> m;
for(int i = 1; i <= n; i++) cin >> a[i];
sort(a+1, a+1+n, [](LL x, LL y) {
return x > y;
});
int tot1 = 0, tot2 = 0;
for(int i = 1; i <= n; i++) {
if(a[i] <= m) A[++tot1] = a[i];
else B[++tot2] = a[i];
}
if(tot1 == 0) {
LL nn = (n+d) / (d + 1);
LL ans = 0;
for(int i = 1; i <= nn; i++) {
ans += B[i];
}
cout << ans << "
";
return 0;
}
if(tot2 == 0) {
LL ans = 0;
for(int i = 1; i <= n; i++) {
ans += A[i];
}
cout << ans << "
";
return 0;
}
for(int i = 1; i <= tot1; i++) sa[i] = sa[i-1] + A[i];
for(int i = 1; i <= tot2; i++) sb[i] = sb[i-1] + B[i];
LL ans = 0;
for(int i = 0; i <= tot1; i++) {
LL cur = sa[i];
LL nn = (n-i+d) / (d+1);
ans = max(cur+sb[nn], ans);
}
cout << ans << "
";
return 0;
}
E
题意
给一个图(G=(V,E)),选择一个tuple ((c_1,dots, c_k))作为遍历图的规则，出度为(i)的点的下一个节点是第(c_i)小的边指向的点，问有多少种tuple可以把图遍历一遍
Solution
- 不能发现其实最后的路径是一个环，可以对每个点的下一个节点所组成的集合是(V)，因此可以通判断过这是否成立来检查一个tuple
- 另外可以将点按出度分类，维护(S[i][j])表示出度为(i)的点，选第(j)小的边所到达的点的集合
- 可以用hash判断集合是否相等，有可能会冲突
#include <cstdio>
#include <stack>
#include <set>
#include <cmath>
#include <map>
#include <time.h>
#include <vector>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
//#include <memory.h>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <cassert>
#include <unordered_map>
#define P pair<int, int>
#define LL long long
#define LD long double
#define PLL pair<LL, LL>
#define mset(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i < b; i++)
#define PI acos(-1.0)
#define random(x) rand() % x
#define debug(x) cout << #x << " " << x << "
"
using namespace std;
const int inf = 0x3f3f3f3f;
const LL __64inf = 0x3f3f3f3f3f3f3f3f;
#ifdef DEBUG
const int MAX = 2e3 + 50;
#else
const int MAX = 2e6 + 500;
#endif
const int mod = 1e9 + 7;
void file_read() {
#ifdef DEBUG
freopen("in", "r", stdin);
// freopen("out", "w", stdout);
#endif
}
#define ULL unsigned long long
const int mod2 = 998244353;
vector<P> G[MAX];
ULL s[100][100];
LL g[100][100];
LL t[100][100];
bool vis[MAX];
ULL val[MAX];
ULL TARGET1 = 1;
LL TARGET2 = 1;
LL TARGET3 = 1;
int n, m, k;
int ans;
void dfs(int u) {
vis[u] = 1;
for(int i = 0; i < G[u].size(); i++) {
int v = G[u][i].first;
s[G[u].size()][i+1] *= (ULL)val[v];
g[G[u].size()][i+1] = g[G[u].size()][i+1] * val[v] % mod;
t[G[u].size()][i+1] = t[G[u].size()][i+1] * val[v] % mod2;
if(vis[v]) continue;
dfs(v);
}
}
void dfs2(int x, ULL cur1, LL cur2, LL cur3) {
if(x > k) {
if(cur1 == TARGET1 and cur2 == TARGET2 and cur3 == TARGET3) ans ++;
return ;
}
for(int i = 1; i <= x; i++) {
dfs2(x+1, cur1*s[x][i], cur2*g[x][i]%mod, cur3 * t[x][i] % mod2);
}
}
int main() {
srand((int)time(0));
file_read();
for(int i = 0; i < 100; i++) for(int j = 0; j < 100; j++) s[i][j] = g[i][j] = t[i][j] = 1;
scanf("%d%d%d", &n, &m, &k);
for(int i = 1; i <= n; i++) val[i] = rand() % mod;
for(ULL i = 1; i <= n; i++) TARGET1 = TARGET1 * val[i], TARGET2 = TARGET2 * val[i] % mod, TARGET3 = TARGET3 * val[i] % mod2;
for(int i = 1; i <= m; i++) {
int u, v, w;
scanf("%d%d%d", &u, &v, &w);
G[u].push_back({v, w});
}
for(int i = 1; i <= n; i++) {
sort(G[i].begin(), G[i].end(), [](const P&A, const P&B){
return A.second < B.second;
});
}
dfs(1);
dfs2(1, 1, 1, 1);
printf("%d
", ans);
return 0;
}
F
题意
定义串只有N和B，有四种操作，给(s_1, dots, s_n)，求一个串(t)使得和(s)的最大距离最小
Solution
- 不难发现字符串对应一个坐标，对于一个(x,y)可以移动上下左右，或者右上，左下
- 对于一个点，在固定距离内可以访问的点是一个六边形
- 要找一个最小半径的六边形覆盖所有的点
- 因此二分半径，计算六边形中心的可行区域
- 最后在可行区域选一个作为(t),不能是空串
#include <cstdio>
#include <stack>
#include <set>
#include <cmath>
#include <map>
#include <time.h>
#include <vector>
#include <iostream>
#include <string>
#include <cstring>
#include <algorithm>
//#include <memory.h>
#include <cstdlib>
#include <queue>
#include <iomanip>
#include <cassert>
#include <unordered_map>
#define P pair<int, int>
#define LL long long
#define LD long double
#define PLL pair<LL, LL>
#define mset(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for (int i = a; i < b; i++)
#define PI acos(-1.0)
#define random(x) rand() % x
#define debug(x) cout << #x << " " << x << "
"
using namespace std;
const int inf = 0x3f3f3f3f;
const LL __inf64 = 0x3f3f3f3f3f3f3f3f;
#ifdef DEBUG
const int MAX = 2e3 + 50;
#else
const int MAX = 2e6 + 500;
#endif
const int mod = 1e9 + 7;
void file_read() {
#ifdef DEBUG
freopen("in", "r", stdin);
// freopen("out", "w", stdout);
#endif
}
vector<P> pt;
int xL, xR, yL, yR, zL, zR;
bool check(int mid) {
xL = -inf, xR = inf, yL = -inf, yR = inf, zL = -inf, zR = inf;
for(auto p : pt) {
xL = max(xL, p.first-mid);
xR = min(xR, p.first+mid);
yL = max(yL, p.second-mid);
yR = min(yR, p.second+mid);
zL = max(zL, p.first-p.second-mid);
zR = min(zR, p.first-p.second+mid);
}
if(xL > xR || yL > yR || zL > zR) return false;
yL = max(yL, xL-zR);
yR = min(yR, xR-zL);
if(yL > yR) return false;
return true;
}
int main() {
file_read();
int n;
cin >> n;
for(int i = 1; i <= n; i++) {
string s;
cin >> s;
int x = 0, y = 0;
for(auto c : s) {
if(c == 'B') x++;
else y++;
}
pt.emplace_back(x, y);
}
int l = 0, r = inf;
while(l < r) {
int mid = (l + r) >> 1;
if(check(mid)) r = mid;
else l = mid + 1;
}
// cout << check(l) << "
";
check(l);
int ansx, ansy;
for(int x = max(0, xL); x <= xR; x++) {
int curyL = max(yR, x-zR);
int curyR = min(yR, x-zL);
if(curyL > curyR) continue;
if(x > 0 || curyR > 0) {
ansx = x;
ansy = curyR;
break;
}
}
cout << l << "
";
while(ansx--) printf("B");
while(ansy--) printf("N");
return 0;
}