CodeForces 1252E Songwriter（贪心）

题目：

Andi is a mathematician, a computer scientist, and a songwriter. After spending so much time writing songs, he finally writes a catchy melody that he thought as his best creation. However, the singer who will sing the song/melody has a unique vocal range, thus, an adjustment may be needed.

A melody is defined as a sequence of $N$

If $A_{i} < A_{i + 1}$
If $A_{i} = A_{i + 1}$
If $A_{i} > A_{i + 1}$
$| B_{i} - B_{i + 1} | \leq K$

Moreover, the singer also requires that all notes are within her vocal range, i.e. $L \leq B_{i} \leq R$

Help Andi to determine whether such $B$

For example, consider a melody $A = {1, 3, 5, 6, 7, 8, 9, 10, 3, 7, 8, 9, 10, 11, 12, 12}$

$\text{[math]}$

Supposed we want to make a new melody with $L = 1$

Input

Input begins with a line containing four integers: $N$

Output

Output in a line $N$

Examples

Input

16 1 8 6
1 3 5 6 7 8 9 10 3 7 8 9 10 11 12 12

Output

1 2 3 4 5 6 7 8 2 3 4 5 6 7 8 8

Input

16 1 8 6
1 3 5 6 7 8 9 10 3 7 8 9 10 11 12 13

Output

-1

Input

16 1 10 10
1 3 5 6 7 8 9 10 3 7 8 9 1 11 12 13

Output

1 2 3 4 5 6 7 8 1 2 3 4 1 2 3 4

Note

Explanation for the sample input/output #1

This is the example from the problem description.

题意：

　　给出长度为n的序列a，要求构造长度为n的序列b且字典序最小，要求相邻的两数之间，b[i]和b[i+1]的大小关系与a[i]和a[i+1]的大小关系相同，并且相邻的abs(b[i]-b[i+1])<=k,同时任意b[i]要在[l,r]的范围内。

思路：

　　因为要求字典序最小，所以我们可以先预处理出每个位置能取的数字范围，然后贪心地从左往右取。

　　预处理数字范围的时候从后往前，分3种情况：

　　1.a[i] == a[i+1]

　　　　这种时候显然l[i]=l[i+1],r[i]=r[i+1]

　　2.a[i] > a[i+1]

　　　　因为梯度最大是k，故r[i] = max(r[i+1]+k,R),在后一个最大可能的基础上再加上k，同时保证范围不越界

　　　　l[i] = l[i+1] + 1,最小的情况只能是最小可能数+1，这时候应判一下越界问题

　　3.a[i] < a[i+1]

　　　　同2，l[i] = max(L,l[i+1] - k)，最小情况是最小可能数-k，同时保证范围不越界

　　　　　　 r[i] = r[i+1] - 1，最大情况是最大可能数-1，判越界问题

代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e5 + 7;
int l[maxn],r[maxn],a[maxn],L,R,n,k;
int main() {
    scanf("%d%d%d%d",&n,&L,&R,&k);
    for (int i=1; i<=n; ++i) {
        scanf("%d",&a[i]);
    }
    int f = 1;
    l[n] = L; r[n] = R;
    for (int i=n-1; i>=1; --i) {
        if(a[i] == a[i+1]) {
            l[i] = l[i+1];
            r[i] = r[i+1];
        }
        else if(a[i] > a[i+1]) {
            l[i] = l[i+1] + 1;
            r[i] = min(r[i+1] + k,R);
        }
        else {
            l[i] = max(L,l[i+1] - k);
            r[i] = r[i+1] - 1;
        }    
        if(l[i] > r[i] || l[i] > R || r[i] < L) f = 0;
    }
    if(f == 0) {
        printf("-1");
        return 0;
    }
    int ans = l[1];
    for (int i=1; i<=n; ++i) {
        printf("%d ",ans);
        if(a[i] == a[i+1]) continue;
        else if(a[i] > a[i+1]) {
            ans = max(ans-k,l[i+1]);
        }
        else ans = max(ans+1,l[i+1]);
    }
    
    return 0;
}

posted @ 2021-03-11 20:50 丿不落良辰阅读(60) 评论(0) 编辑收藏举报

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CodeForces 1252E Songwriter（贪心）

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