Educational Codeforces Round 62 (Div. 2)
自测的时候做出了A~E
E花的时间太多q-q
A Detective Book
暴力模拟
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int N = 1e4 + 5;
int n, ans, pre;
int main(){
scanf("%d", &n);
for(int i = 1, x; i <= n; ++i){
scanf("%d", &x);
pre = max(x, pre);
if(i >= pre){++ans; pre = 0;}
}
printf("%d", ans);
return 0;
}
B Good String
取最左边的">"和最右边的"<"与最左/右边距离的最小值
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int N = 105;
int n, ans, fir, las;
char str[N];
int main(){
int T; scanf("%d", &T);
while(T--){
scanf("%d%s", &n, str + 1);
fir = n, las = 1;
for(int i = 1; i <= n; ++i) if(str[i] == '>') {fir = i; break;}
for(int i = n; i >= 1; --i) if(str[i] == '<') {las = i; break;}
ans = min(fir - 1, n - las);
printf("%d\n", ans);
}
return 0;
}
C Playlist
按bi从大到小
用multiset模拟1~i - 1中t最大的k个的和
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <set>
using namespace std;
const int N = 3e5 + 5;
int n, m;
long long ans;
struct Node{
int x, y;
}node[N];
inline bool rule(Node x, Node y){
return x.y > y.y;
}
multiset<int> st;
long long stsize, len;
int main(){
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i){
scanf("%d%d", &node[i].x, &node[i].y);
}
sort(node + 1, node + n + 1, rule);
for(int i = 1; i <= n; ++i){
if(st.size() < m){
st.insert(node[i].x);
len += node[i].x;
} else if((*st.begin()) < node[i].x){
len += node[i].x - (*st.begin());
st.erase(st.begin()); st.insert(node[i].x);
}
long long d = 1ll * len * node[i].y;
ans = max(d, ans);
}
printf("%lld", ans);
return 0;
}
D Minimum Triangulation
手玩一下。。。从一放射形划分
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <set>
using namespace std;
const int N = 3e5 + 5;
int n;
long long ans;
int main(){
scanf("%d", &n);
for(int i = 3; i <= n; ++i){
ans += 1ll * i * (i - 1);
}
printf("%lld", ans);
return 0;
}
E Palindrome-less Arrays
有一点麻烦的dp
suf是每次向后跳两个
找到第一个不为-1的数就记录
如果没有 那么记录为后面的最后一个-1 再没有就记为零
f[i][0]是第i个所在-1连续子串在i的位置与suf不同的种类数
f[i][1]是第i个所在-1连续子串在i的位置与suf相同的种类数
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <set>
using namespace std;
const int N = 3e5 + 5;
const long long P = 998244353;
int n, k;
int a[N], suf[N], pre[N];
long long f[N][2], ans = 1;
int main(){
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; ++i){
scanf("%d", &a[i]);
if(~a[i] && a[i] == a[i - 2]){
printf("0"); return 0;
}
}
for(int i = n - 2; i >= 1; --i){
if(~a[i + 2]) suf[i] = a[i + 2];
else suf[i] = suf[i + 2];
}
for(int i = 3; i <= n; ++i) pre[i] = a[i - 2];
for(int i = 1; i <= n; ++i){
if(~a[i]){
if(suf[i] == a[i]){
f[i][1] = 1;
}
else {
f[i][0] = 1;
}
}
else {
if(pre[i]){
if(suf[i]){
if(suf[i] != a[i + 2]) f[i][1] = f[i - 2][0];
f[i][0] = (f[i - 2][0] * (k - 2) % P + f[i - 2][1] * (k - 1) % P) % P;
}
else {
f[i][1] = 0;
f[i][0] = f[i - 2][0] * (k - 1) % P;
}
}
else {
if(suf[i]){
if(suf[i] != a[i + 2]) f[i][1] = 1;
f[i][0] = k - 1;
}
else {
f[i][1] = 0;
f[i][0] = k;
}
}
}
// printf("%d %d %lld %lld\n", pre[i], suf[i], f[i][0], f[i][1]);
}
for(int i = 1; i <= n; ++i){
if(suf[i] == a[i + 2]){
ans = ans * f[i][0] % P;
}
}
printf("%lld\n", ans);
return 0;
}
