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;   
}

F Extending Set of Points

G Double Tree