C - Make a Rectangle
每次取两个相同的且最大的边，取两次即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 100005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int N,a[MAXN];
map<int,int> zz;
void Solve() {
read(N);
for(int i = 1 ; i <= N ; ++i) {read(a[i]);zz[a[i]]++;}
int r = 0,c = 0;
while(1) {
if(zz.empty()) break;
auto t = *(--zz.end());zz.erase(--zz.end());
if(t.se >= 2) {
if(!r) r = t.fi;
else if(!c) c = t.fi;
}
if(t.se > 2) zz[t.fi] = t.se - 2;
if(r && c) break;
}
out(1LL * r * c);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
D - Coloring Dominoes
就是如果是
竖条对两个长条，那么方案数乘上2
竖条对竖条，乘上2
两个长条+竖条，方案数乘上1
两个长条+两个长条 有3种
1 2
2 1
和
1 3
2 1
和
1 2
2 3
初始如果一个竖条有3种，两个横条有6种
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 100005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N,f[65];
char s[2][65];
int inc(int a,int b) {
return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
return 1LL * a * b % MOD;
}
void Solve() {
read(N);
scanf("%s%s",s[0] + 1,s[1] + 1);
int ans = 0;
for(int i = 1 ; i <= N ; ++i) {
if(s[0][i] == s[0][i + 1]) continue;
if(s[0][i - 1] == s[0][i]) {
if(i <= 2) ans = 6;
else {
if(f[i - 2] == 0) ans = mul(ans,3);
else if(f[i - 2] == 1) ans = mul(ans,2);
}
f[i] = 0;
}
else {
if(i <= 1) ans = 3;
else {
if(f[i - 1] == 1) ans = mul(ans,2);
}
f[i] = 1;
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
E - Don't Be a Subsequence
撞题了吧，最短不公共子串的那个模板大礼包
直接建序列自动机，求最短路，每次从前往后遍历走一条最短路边
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 200005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
char a[MAXN];
int nxt[MAXN][26],N,dis[MAXN];
string ans = "";
void Solve() {
scanf("%s",a + 1);
N = strlen(a + 1);
for(int i = 0 ; i < 26 ; ++i) nxt[N][i] = N + 1;
for(int i = N - 1 ; i >= 0 ; --i) {
for(int j = 0 ; j < 26 ; ++j) nxt[i][j] = nxt[i + 1][j];
nxt[i][a[i + 1] - 'a'] = i + 1;
}
dis[N + 1] = 0;
for(int i = N ; i >= 0 ; --i) {
dis[i] = N + 1;
for(int j = 0 ; j < 26 ; ++j) dis[i] = min(dis[nxt[i][j]] + 1,dis[i]);
}
int pos = 0;
while(pos != N + 1) {
for(int i = 0 ; i < 26 ; ++i) {
if(dis[nxt[pos][i]] + 1 == dis[pos]) {
ans += (i + 'a');
pos = nxt[pos][i];
break;
}
}
}
cout << ans << endl;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}
F - Flip and Rectangles
我们认为一个长方形任意一个22的方格如果包含偶数个黑点
那么一定会变成全黑
我们把每个22的方格中间标记成一个新点，判断能否选择，然后用最大子矩形的算法做
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define MAXN 2005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
res = 0;T f = 1;char c = getchar();
while(c < '0' || c > '9') {
if(c == '-') f = -1;
c = getchar();
}
while(c >= '0' && c <= '9') {
res = res * 10 + c - '0';
c = getchar();
}
res *= f;
}
template<class T>
void out(T x) {
if(x < 0) {x = -x;putchar('-');}
if(x >= 10) {
out(x / 10);
}
putchar('0' + x % 10);
}
int H,W,ans;
char s[MAXN][MAXN];
int tr[MAXN],tl[MAXN],h[MAXN],l[MAXN],r[MAXN];
bool check(int x,int y) {
return ((s[x - 1][y - 1] == '#') + (s[x - 1][y] == '#') + (s[x][y - 1] == '#') + (s[x][y] == '#')) & 1;
}
void Solve() {
read(H);read(W);
ans = max(H,W);
for(int i = 0 ; i < H ; ++i) {
scanf("%s",s[i]);
}
for(int i = 1 ; i <= W - 1; ++i) l[i] = 0,r[i] = W;
tr[W] = W;
for(int i = 1 ; i < H ; ++i) {
for(int j = 1 ; j < W ; ++j) {
if(check(i,j)) tl[j] = j;
else tl[j] = tl[j - 1];
}
for(int j = W - 1 ; j >= 1 ; --j) {
if(check(i,j)) tr[j] = j;
else tr[j] = tr[j + 1];
}
for(int j = 1 ; j < W ; ++j) {
if(check(i,j)) {
h[j] = 0,l[j] = 0,r[j] = W;
}
else {
l[j] = max(tl[j],l[j]);
r[j] = min(tr[j],r[j]);
h[j]++;
ans = max(ans,(r[j] - l[j] - 1 + 1) * (h[j] + 1));
}
}
}
out(ans);enter;
}
int main() {
#ifdef ivorysi
freopen("f1.in","r",stdin);
#endif
Solve();
}