序列 [构造题]

$序列$

---

$color{red}{正解部分}$

首先 最多有一个数字同时出现在 最长上升子序列 和 最长下降子序列 中, 所以若 $A+B > N+1$, 说明 无解 .

然后考虑怎么构造, 将 ${1,2,3...N}$ 分成若干个大小为 $B$ 的块(可能会剩下一个小块), 总共有 $a = lceil frac{N}{B} ceil$ 个块,

则 最长上升子序列 的长度至少为 $a$, 若 $A < a$, 说明 无解 .
否则可以在 块 内反转元素, 每反转一次 最长上升子序列 的长度会增加 $1$, 最长下降子序列 的长度不变 .

按上述方法构造即可 .

---

$color{red}{实现部分}$

$将剩余部分放在前面$

#include<bits/stdc++.h>
#define reg register

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 1e5 + 10;

int N;
int A;
int B;

bool used[maxn];

void Work(){
        N = read(), A = read(), B = read();
        for(reg int i = 1; i <= N; i ++) used[i] = 0;
        int left = N / B;
        if(N % B) left ++;
        if(A+B > N+1 || A < left){ printf("No
"); return ; }
        printf("Yes
");
        int base = N % B;      
        for(reg int i = 1; i <= A-left; i ++){
                used[i] = 1, printf("%d ", i);
                if((i - base) % B == 0) A ++;
        }
        int t = N%B;
        while(t <= N){
                int t2 = std::max(1, t - B + 1), tmp = t;
                while(tmp >= t2){
                        if(!used[tmp]) printf("%d ", tmp);
                        tmp --;
                }
                t += B;
        }
        printf("
");
        /*
        if(t >= N + 1){
                for(reg int j = 0 ; j < t; j ++) printf("%d ", ++ cur);
        }
        for(reg int i = 1; i <= B; i ++) printf("%d ", i-1);
        */
}

int main(){
        int T = read();
        while(T --) Work();
        return 0;
}

---

$将剩余部分放在后面$

#include<bits/stdc++.h>
#define reg register

int read(){
        char c;
        int s = 0, flag = 1;
        while((c=getchar()) && !isdigit(c))
                if(c == '-'){ flag = -1, c = getchar(); break ; }
        while(isdigit(c)) s = s*10 + c-'0', c = getchar();
        return s * flag;
}

const int maxn = 1e6 + 10;

int N;
int A;
int B;

bool used[maxn];

void Work(){
        N = read(), A = read(), B = read();
        for(reg int i = 1; i <= N; i ++) used[i] = 0;
        int left = N / B;
        if(N % B) left ++;
        if(A+B > N+1 || A < left){ printf("No
"); return ; }
        printf("Yes
");
        for(reg int i = 1; i <= A-left; i ++){
                used[i] = 1, printf("%d ", i);
                if(i % B == 0) A ++;
        }
        int t = B;
        while(t <= N){
                int t2 = t - B + 1, tmp = t;
                while(tmp >= t2){
                        if(!used[tmp]) printf("%d ", tmp);
                        tmp --;
                }
                t += B;
        } 
        for(reg int i = N; i >= N-(N%B)+1; i --) if(!used[i]) printf("%d ", i);
        printf("
");
}

int main(){
        freopen("seq.in", "r", stdin);
        freopen("seq.out", "w", stdout);
        int T = read();
        while(T --) Work();
        return 0;
}