time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
We have a secret array. You don't know this array and you have to restore it. However, you know some facts about this array:
The array consists of nn distinct positive (greater than 00) integers.The array contains two elements xx and yy (these elements are known for you) such that x<yx<y.If you sort the array in increasing order (such that a1<a2<…<ana1<a2<…<an), differences between all adjacent (consecutive) elements are equal (i.e. a2−a1=a3−a2=…=an−an−1)a2−a1=a3−a2=…=an−an−1).It can be proven that such an array always exists under the constraints given below.
Among all possible arrays that satisfy the given conditions, we ask you to restore one which has the minimum possible maximum element. In other words, you have to minimize max(a1,a2,…,an)max(a1,a2,…,an).
You have to answer tt independent test cases.
Input
The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases. Then tt test cases follow.
The only line of the test case contains three integers nn, xx and yy (2≤n≤502≤n≤50; 1≤x<y≤501≤x<y≤50) — the length of the array and two elements that are present in the array, respectively.
Output
For each test case, print the answer: nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109), where aiai is the ii-th element of the required array. If there are several answers, you can print any (it also means that the order of elements doesn't matter).
It can be proven that such an array always exists under the given constraints.
Example
input
Copy
5 2 1 49 5 20 50 6 20 50 5 3 8 9 13 22output
Copy
1 49 20 40 30 50 10 26 32 20 38 44 50 8 23 18 13 3 1 10 13 4 19 22 25 16 7
解题说明：此题是一道数学题，按照题目意思可以发现数字最多为50个，于是可以直接从小到大枚举x~y之间的间距，看看最多能放几个数，再把剩下的数从 x往前放，多了再从y往后放。
#include <stdio.h> int main() { long long int t, n, x, y, ans; scanf("%lld", &t); while (t--) { scanf("%lld %lld %lld", &n, &x, &y); int d = y - x; int diff; for (int i = 1; i <= d; i++) { if (d%i == 0) { diff = i; if (y - (n - 1)*i <= x) { break; } } } int start = y - (n - 1)*diff; while (start <= 0) { start += diff; } for (int i = 0; i<n; i++) { printf("%d ", start + (diff*i)); } printf("\n"); } return 0; }
