CF1739 部分题解
比赛链接：https://codeforces.com/contest/1739
AB
水题
// by SkyRainWind
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mpr make_pair
#define debug() cerr<<"Yoshino\n"
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int inf = 1e9, INF = 0x3f3f3f3f;
signed main(){
int te;scanf("%d",&te);
while(te --){
int n,m;scanf("%d%d",&n,&m);
if(n == 1 || m == 1){
puts("1 1");continue;
}
if(max(n, m) >= 4){
puts("1 1");continue;
}
puts("2 2");
}
return 0;
}
// by SkyRainWind
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mpr make_pair
#define debug() cerr<<"Yoshino\n"
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int inf = 1e9, INF = 0x3f3f3f3f;
void solve(){
vector<int>vc;vc.clear();
int n;scanf("%d",&n);
int gg=0;
for(int i=1;i<=n;i++){
int x;scanf("%d",&x);
if(i == 1)vc.push_back(x);
else{
int lst = vc[vc.size() - 1];
if(x > 0 && lst - x >= 0){
gg=1;
}else vc.push_back(x + lst);
}
}
if(gg){puts("-1");return ;}
for(int i : vc)printf("%d ",i);puts("");
}
signed main(){
int te;scanf("%d",&te);
while(te --)solve();
return 0;
}
C
以 n=8为例，考虑draw的情况就是 2 3 6 7 || 1 4 5 8这种“螺旋”的，从大到小每次枚举当前剩下最大值在哪，每次加一个组合数即可，判一下最后一个1在哪
// by SkyRainWind
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mpr make_pair
#define debug() cerr<<"Yoshino\n"
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int inf = 1e9, INF = 0x3f3f3f3f, mod =998244353;
int fac[65], inv[65];
int pw(int x,int y){
if(y == 0)return 1;
if(y == 1)return x;
int mid = pw(x, y>>1);
if(y&1)return 1ll*mid*mid%mod*x%mod;
return 1ll*mid*mid%mod;
}
int C(int x,int y){
return 1ll * fac[x] * inv[x-y] % mod * inv[y]%mod;
}
void solve(){
int n;scanf("%d",&n);
int r1 = 0, r2 = 0;
int cur = n -1 ;
for(int i = n/2 -1,p=1; i>=1; i--,p++){
if(i == n/2-1)r1 += C(cur,i), --cur;
else{
if(p&1)r1 += (C(cur, i)+C(cur-1,i))%mod;
else r2 += (C(cur,i)+C(cur-1,i))%mod;
cur -= 2;
}
r1 %= mod, r2 %= mod;
}
if(n%4 == 0)++ r2;
else ++ r1;
if(n%4 == 0)++ r2;
else if(n>2)++ r1;
printf("%d %d 1\n",r1,r2);
}
signed main(){
fac[0] = inv[0] = 1;
for(int i=1;i<=60;i++)fac[i] = 1ll*fac[i-1]*i%mod;
inv[60] = pw(fac[60], mod -2 );
for(int i=59;i>=1;i--)inv[i] = 1ll * inv[i+1]*(i+1)%mod;
int te;scanf("%d",&te);
while(te --)solve();
return 0;
}
D
显然先二分答案
直接贪心会有反例，比如：
1 5 1 1 2 3 3
答案是2而非3，断2->3边
这启示我们可以自底向上贪心，每次遇到当前子树最大深度>=mid的话而且当前点不是根且父亲不是根，就把当前子树连到根
// by SkyRainWind
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mpr make_pair
#define debug() cerr<<"Yoshino\n"
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int inf = 1e9, INF = 0x3f3f3f3f, maxn = 2e5 + 5;
int n,k;
vector<int>g[maxn];
int dep[maxn];
int tot = 0;
int dfs2(int x,int fat,int lim){
int curd = 0;
for(int u : g[x]){
curd = max(curd, dfs2(u, x,lim ));
}
++ curd;
if(fat != 1 && x != 1 && curd >= lim){++tot;return 0;}
return curd;
}
int check(int x){
tot = 0;
dfs2(1,-1,x);
// printf("?? %d\n",tot);
if(tot > k)return 0;
return 1;
}
void solve(){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)g[i].clear();
for(int i=2;i<=n;i++){
int p;scanf("%d",&p);
g[p].push_back(i);
}
int l=1,r=n, ans;
// for(int i=1;i<=n;i++)printf("%d ",dep[i]);puts("");
// printf("%d\n",check(1));
// return ;
while(l <= r){
int mid = l +r >>1;
if(check(mid))r = mid-1, ans =mid;
else l = mid+1;
}
printf("%d\n",ans);
}
signed main(){
int te;scanf("%d",&te);
while(te --)solve();
return 0;
}
E
手玩一下全是1的样例发现，如果会出现二选一的情况，一定是(i,j)->(i,j+1)->(i+1,j+1)->(i+1,j+2)这种路径（其中(i,j+1)是1跳过了）
考虑dp，设\(dp[i][j]\)表示到(i,j)结尾的答案，其中是第一次到第j列
转移考虑普通的向右转移，还有一种如果(i^1,j)的数为1的话，那么(i,j+1)如果也为1，就是上面的螺旋走最优，如果为0，显然也是经过1更优
分析一下似乎情况就全了
// by SkyRainWind
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
#define mpr make_pair
#define debug() cerr<<"Yoshino\n"
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)
#define pii pair<int,int>
using namespace std;
typedef long long LL;
const int inf = 1e9, INF = 0x3f3f3f3f, maxn = 2e5+5;
int n;
char s[2][maxn];
int dp[2][maxn][2];
void ck(int &mm,int a){mm = max(mm, a);}
signed main(){
memset(dp, -0x3f, sizeof dp);
scanf("%d",&n);
for(int i=0;i<=1;i++)scanf("%s", s[i] + 1);
s[0][n+1] = s[1][n+1] = '0';
// dp[0][0][0] = 0;
// printf("%d\n",dp[0][0][0]);
dp[0][1][s[1][1] - '0'] = s[1][1] - '0';
for(int i=1;i<=n;i++){
for(int j=0;j<=1;j++){
}
}
// printf("%d\n",dp[1][2][1]);
printf("%d\n",max(dp[0][n+1][0], dp[1][n+1][0]));
return 0;
}
