cf div2 round 688 题解
作者：互联网
爆零了，自闭了
小张做项目入职字节
小李ak wf入职ms
我比赛爆零月薪3k
我们都有光明的前途
好吧，这场感觉有一点难了，昨天差点卡死在B上，要不受O爷出手相救我就boom zero了
第一题，看上去很唬人，我觉得还得记录变量什么的，1分钟后发现只要查出两个集合的交集大小就行了，连变量增量都不用，拿一血，此时rk 1k6，我人没了
#include <bits/stdc++.h>
using namespace std;
#define limit (315000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
ll sign = 1, x = 0;char s = getchar();
while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
return x * sign;
}//快读
void write(ll x){
if(x < 0) putchar('-'),x = -x;
if(x / 10) write(x / 10);
putchar(x % 10 + '0');
}
int kase;
int n,m,k;
int a[limit];
void solve(){
cin>>n>>m;
set<int>s;
rep(i,1,n){
int x;
cin>>x;
s.insert(x);
}
int tot = 0;
rep(i,1,m){
int x;
cin>>x;
if(s.find(x) != s.end())++tot;
}
cout<<tot<<endl;
}
int main() {
#ifdef LOCAL
FOPEN;
#endif
cin>>kase;
while (kase--){
solve();
}
return 0;
}
B题，我现在其实没太搞懂这个怎么搞的，挖坑
#include <bits/stdc++.h>
using namespace std;
#define limit (3150000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
ll sign = 1, x = 0;char s = getchar();
while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
return x * sign;
}//快读
void write(ll x){
if(x < 0) putchar('-'),x = -x;
if(x / 10) write(x / 10);
putchar(x % 10 + '0');
}
int kase;
int n,m,k;
ll a[limit];
ll dp2[limit],dp[limit];
void solve(){
cin>>n;
ll tot = 0,ans = 0x3f3f3f3f3f3f3f;
rep(i,1,n){
cin>>a[i];
dp[i] = 0;
dp2[i] = 0;
}
dp2[1] = max(dp2[1],llabs(a[2] - a[1]));
rep(i,2,n){
ll tmp = llabs(a[i] - a[i - 1]);
tot += tmp;
dp[i] = tmp;
}
rep(i,1,n){
if(i == 1){
ans = min(ans , tot - dp[2]);
}else if(i == n){
ans = min(ans,tot - dp[n]);
}
else if((a[i] <= a[i + 1] && a[i] <= a[i - 1])){
ans = min(ans,tot + abs(a[i - 1] - a[i + 1]) - dp[i + 1 ] - dp[i]);
}else if(a[i] >= a[i + 1] && a[i] >= a[i - 1]){
ans = min(ans,tot + abs(a[i + 1] - a[i - 1]) - dp[i + 1] - dp[i]);
}
}
cout<<ans<<endl;
}
int main() {
#ifdef LOCAL
FOPEN;
#endif
cin>>kase;
while (kase--){
solve();
}
return 0;
}
C这个思路很显然，就是看每次（1）是否能选取到两个端点，然后贪心去放第三个顶点，因为我们可以改变一个格子，所以肯定要顶着1或者n放
（2）我们只用一个端点到两边，看哪个底大，是否能构成更大的三角形，顶点是否存在
（3）同上，但这次只有一个顶点
（4）这一行没顶点，拉闸
极不好写，100+的大暴力，给人写自闭了，不过好在1A，写完一身汗
#include <bits/stdc++.h>
using namespace std;
#define limit (315000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
ll sign = 1, x = 0;char s = getchar();
while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
return x * sign;
}//快读
void write(ll x){
if(x < 0) putchar('-'),x = -x;
if(x / 10) write(x / 10);
putchar(x % 10 + '0');
}
int kase;
int n,m,k;
ll a[limit];
int mp[2005][2006];
int l[2005][11],r[2005][11],up[2005][11],down[2006][11];
void solve(){
cin>>n;
rep(i,0,9)a[i] = 0;
rep(i,1,n){
string str;
cin>>str;
rep(h,0,9)l[i][h] = r[i][h] =down[i][h] =up[i][h] =0;
rep(j,1,n){
mp[i][j] = str[j-1] - '0';
}
}
rep(i,1,n){
rep(p,0,9){
rep(j,1,n){
if(mp[i][j] == p){
r[i][p] = j;
break;
}
}
per(j,1,n){
if(mp[i][j] == p){
l[i][p] = j;
break;
}
}
}
}
rep(j,1,n){
rep(p,0,9){
rep(i,1,n){
if(mp[i][j] == p){
down[j][p] = i;
break;
}
}
per(i,1,n){
if(mp[i][j] == p){
up[j][p] = i;
break;
}
}
}
}
rep(p,0,9){
rep(i,1,n){
if(l[i][p] != r[i][p]){
int base = abs(l[i][p] - r[i][p]);
a[p] = max({a[p],base * abs(1 - i), base * abs(n - i)});
base = max(abs(l[i][p] - 1), abs(l[i][p] - n));
rep(j,1,n){
if(up[j][p])a[p] = max({a[p], base * abs(up[j][p] - i), base* abs(down[j][p] - i)});
}
base = max(abs(r[i][p] - 1), abs(r[i][p] - n));
rep(j,1,n){
if(up[j][p])a[p] = max({a[p], base * abs(up[j][p] - i), base* abs(down[j][p] - i)});
}
}else if(l[i][p]){
int base = max(abs(1 - l[i][p]),abs(l[i][p] - n));
rep(j,1,n){
if(up[j][p])a[p] = max({a[p], base * abs(up[j][p]- i),base * abs(down[j][p] - i)});
}
}
if(up[i][p] != down[i][p]){
int base = abs(up[i][p] - down[i][p]);
a[p] = max({a[p],base * abs(1 - i), base * abs(n - i)});
base = max(abs(up[i][p] - 1), abs(up[i][p] - n));
rep(j,1,n){
if(l[j][p])a[p] = max({a[p],base * abs(l[j][p] - i), base * abs(r[j][p] - i)});
}
base = max(abs(down[i][p] - 1), abs(down[i][p] - n));
rep(j,1,n){
if(l[j][p])a[p] = max({a[p],base * abs(l[j][p] - i), base * abs(r[j][p] - i)});
}
}else if(up[i][p]){
int base = max(abs(1 - up[i][p]),abs(down[i][p] - n));
rep(j,1,n){
if(l[j][p])a[p] = max({a[p], base * abs(l[j][p]- i),base * abs(r[j][p] - i)});
}
}
}
}
rep(i,0,9){
cout<<a[i]<<' ';
}
cout<<endl;
}
int main() {
#ifdef LOCAL
FOPEN;
#endif
cin>>kase;
while (kase--){
solve();
}
return 0;
}
D题没时间搞了，但后来和Dxtst老哥讨论了下似乎是当前位上0或1，如果是1对于答案的贡献为2^k, 其他结论今晚再搞
奥里给！
标签：题解,rep,cf,up,abs,688,base,max,define 来源： https://www.cnblogs.com/tiany7/p/14090229.html