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 sequencec = [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.
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.
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.
5 1 5
1 1 1 1 1
3 1 5 4 2
3 1 5 4 2
4 2 9
3 4 8 9
3 2 1 4
2 2 2 9
6 1 5
1 1 1 1 1 1
2 3 5 4 1 6
-1
Sequence b which was found in the second sample is suitable, because calculated sequencec = [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].
题意：
给你序列p和序列a，让你找一个合法的序列b，输出。
序列c是通过bi-ai来得到的。
序列p是通过离散化序列c来得到的。（离散化c序列，就是将c序列中的数字从小到大编号）
#include <iostream> #include<cstdio> #include<algorithm> using namespace std; struct node { int id,num; }a[100005]; int b[100005],k[100005]; int n,l,r; bool cmp(node a,node b) { return a.id>b.id; } int main() { scanf("%d%d%d",&n,&l,&r); for(int i=1;i<=n;i++) scanf("%d",&a[i].num); for(int i=1;i<=n;i++) {scanf("%d",&a[i].id); k[i]=a[i].id;} sort(a+1,a+n+1,cmp); b[1]=r; bool flag=1; for(int i=2;i<=n;i++) { /* if (b[i-1]-a[i-1].num-1+a[i].num<l || b[i-1]-a[i-1].num-1+a[i].num>r) { flag=0; break; } b[i]=b[i-1]-a[i-1].num-1+a[i].num; //以上是错误的,只考虑了差是连续，若是不连续没考虑。 */ if (b[i-1]-a[i-1].num-1+a[i].num>=l) { b[i]=b[i-1]-a[i-1].num-1+a[i].num; if (b[i]>r) b[i]=r; //使差减小得更多。 } else { flag=0; break; } } if (!flag) printf("-1"); else{ for(int i=1;i<=n;i++) { if (i-1) printf(" "); printf("%d",b[n-k[i]+1]); } } printf(" "); return 0; }