2017 Multi-University Training Contest
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Inversion
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 218 Accepted Submission(s): 148
Total Submission(s): 218 Accepted Submission(s): 148
Problem Description
Give an array A, the index starts from 1.
Now we want to knowBi=maxi∤jAj , i≥2 .
Now we want to know
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with one integer n : the size of array A.
Next one line contains n integers, separated by space, ith number isAi .
Limits
T≤20
2≤n≤100000
1≤Ai≤1000000000
∑n≤700000
Each case begins with one line with one integer n : the size of array A.
Next one line contains n integers, separated by space, ith number is
Limits
Output
For each test case output one line contains n-1 integers, separated by space, ith number isBi+1 .
Sample Input
241 2 3 441 4 2 3
Sample Output
3 4 32 4 4
题目描述
首先先普及一下’a|b’代表a是b的因子,即b%a=0,’a∤b’也就是b%a!=0
给定一个序列,求i从2开始不是i倍数的序列中的最大值,依次输出
解题思路
暴力大法好,首先用一个结构体数组保存值与下标。然后对值进行从大到小的排序,然后依此对每个下标i找出对i取余不为零的输出
/* ***********************************************┆ ┏┓ ┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃ ┃ ┆┆┃ ━ ┃ ┆┆┃ ┳┛ ┗┳ ┃ ┆┆┃ ┃ ┆┆┃ ┻ ┃ ┆┆┗━┓ 马 ┏━┛ ┆┆ ┃ 勒 ┃ ┆ ┆ ┃ 戈 ┗━━━┓ ┆┆ ┃ 壁 ┣┓┆┆ ┃ 的草泥马 ┏┛┆┆ ┗┓┓┏━┳┓┏┛ ┆┆ ┃┫┫ ┃┫┫ ┆┆ ┗┻┛ ┗┻┛ ┆************************************************ */#include <iostream>#include <set>#include <map>#include <stack>#include <cmath>#include <queue>#include <cstdio>#include <bitset>#include <string>#include <vector>#include <iomanip>#include <cstring>#include <algorithm>#include <functional>#define inf 0x3f3f3f3f#define debug(x) cout<<"---"<<x<<"---"<<endltypedef long long ll;using namespace std;struct node{ int pos; int val;} a[100500];bool cmp(node x, node y){ return x.val > y.val;}int main(){ int t; cin >> t; while (t--) { int n; cin >> n; for (int i = 1; i <= n; i++) { a[i].pos = i; cin >> a[i].val; } sort(a + 1, a + n + 1, cmp); for (int k = 2; k <= n; k++) { for (int i = 1; i <= n; i++) { if (a[i].pos % k) { if (k == n) { cout << a[i].val << endl; } else { cout << a[i].val << " "; } break; } } } } return 0;}/************************************************┆ ┏┓ ┏┓ ┆┆┏┛┻━━━┛┻┓ ┆┆┃ ┃ ┆┆┃ ━ ┃ ┆┆┃ ┳┛ ┗┳ ┃ ┆┆┃ ┃ ┆┆┃ ┻ ┃ ┆┆┗━┓ ┏━┛ ┆┆ ┃ ┃ ┆ ┆ ┃ ┗━━━┓ ┆┆ ┃ AC代马 ┣┓┆┆ ┃ ┏┛┆┆ ┗┓┓┏━┳┓┏┛ ┆┆ ┃┫┫ ┃┫┫ ┆┆ ┗┻┛ ┗┻┛ ┆************************************************ */
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- 2017 Multi-University Training Contest
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- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
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- 2017 Multi-University Training Contest
- #2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
- 2017 Multi-University Training Contest
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