time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
Dasha logged into the system and began to solve problems. One of them is as follows:
Given two sequences a and b of length n each you need to write a sequence c of length n, the i-th element of which is calculated as follows: ci = bi - ai.
About sequences a and b we know that their elements are in the range from l to r. More formally, elements satisfy the following conditions: l ≤ ai ≤ r and l ≤ bi ≤ r. About sequence c we know that all its elements are distinct.
Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence a and the compressed sequence of the sequence c were known from that test.
Let’s give the definition to a compressed sequence. A compressed sequence of sequence c of length n is a sequence p of length n, so that pi equals to the number of integers which are less than or equal to ci in the sequence c. For example, for the sequence c = [250, 200, 300, 100, 50] the compressed sequence will be p = [4, 3, 5, 2, 1]. Pay attention that in c all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to n inclusively.
Help Dasha to find any sequence b for which the calculated compressed sequence of sequence c is correct.
Input
The first line contains three integers n, l, r (1 ≤ n ≤ 105, 1 ≤ l ≤ r ≤ 109) — the length of the sequence and boundaries of the segment where the elements of sequences a and b are.
The next line contains n integers a1, a2, …, an (l ≤ ai ≤ r) — the elements of the sequence a.
The next line contains n distinct integers p1, p2, …, pn (1 ≤ pi ≤ n) — the compressed sequence of the sequence c.
Output
If there is no the suitable sequence b, then in the only line print “-1”.
Otherwise, in the only line print n integers — the elements of any suitable sequence b.
Examples
input
5 1 5
1 1 1 1 1
3 1 5 4 2
output
3 1 5 4 2
input
4 2 9
3 4 8 9
3 2 1 4
output
2 2 2 9
input
6 1 5
1 1 1 1 1 1
2 3 5 4 1 6
output
-1
Note
Sequence b which was found in the second sample is suitable, because calculated sequence c = [2 - 3, 2 - 4, 2 - 8, 9 - 9] = [ - 1, - 2, - 6, 0] (note that ci = bi - ai) has compressed sequence equals to p = [3, 2, 1, 4].
【题目链接】:http://codeforces.com/contest/761/problem/D
【题解】
因为
c[i] = b[i]-a[i];
题目所给的p[i]实际上是c[i]之间的大小关系;
p[i]=1是最小的,p[i]=2则是第二小…p[i]=n则是最大的;
这样我们可以让
c的值从小到大为
c[1],c[1]+1,c[1]+2,c[1]+3…c[1]+n-1;
即每个值递增1
这样能够在满足p[]数组的情况下尽可能地让c[i]小,相应的b[i]也就小了
但是有可能我们要求c[i]为递增的规律的时候;
b[i]是根据c[i]求出来的
某个i会出现
b[i]< l
这个时候让b[i]为l(b数组没有说一定要全都不同)
然后c[i]相应地调整;
c[i]会变大一点,这样能保证满足p数组,同时b[i]还是最小的l,所以还是能够保证c[i]此时也是最小的
(但是如果b[i]>r,则直接输出无解;
因为此前的操作我们已经尽可能的让c数组小了;
如果还不行(即b[i]还是太大),那么就无解了;)
然后再从i+1开始
让
c[i+1]=c[i]+1
c[i+2] = c[i]+2
…
即此后每个值还是递增1
然后求出相应的b数组
(所以这个c数组最后可能中间会有不是c[i]=c[i-1]+1的断层);
一开始的时候设最小的c的下标为idx
让b[idx]=l;(这样能保证b数组、c数组一开始是最小的了)
求出这个时候的c[idx]作为发生器;
然后做上面的过程就好;
贪心点就是时刻保证c、b数组是最小的,这样b数组在某个时刻通过c数组获得的时候,如果大于r则肯定无解;
小于l则可以调整为l,还是保证c、b为最小的；
【完整代码】
#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
typedef pair<int,int> pii;
typedef pair<LL,LL> pll;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int MAXN = 1e5+100;
int a[MAXN],n,l,r,cc[MAXN],b[MAXN],c[MAXN];
void wujie()
{
puts("-1");
exit(0);
}
int main()
{
//freopen("F:\rush.txt","r",stdin);
cin >> n >> l >> r;
rep1(i,1,n)
rei(a[i]);
rep1(i,1,n)
{
int x;
rei(x);
cc[x] = i;
}
//c[i] = b[i]-a[i];
//c[idx]>c[1]
//b[idx]-a[idx]>b[1]-a[1];
//c[i]+a[i] = b[i];
b[cc[1]] = l;
c[cc[1]] = b[cc[1]]-a[cc[1]];
int temp = b[cc[1]]-a[cc[1]];
rep1(i,2,n)
{
int idx = cc[i];
//
c[idx] = temp+1;
b[idx] = c[idx]+a[idx];
temp++;
if (b[idx]<l)
{
b[idx] = l;
temp = b[idx]-a[idx];
}
if (b[idx]>r)
wujie();
}
for (int i = 1;i <= n;i++)
{
printf("%d",b[i]);
if (i==n)
puts("");
else
putchar(' ');
}
return 0;
}