标签：mis stdio.h rate values integer 区间 minimum elements cin
Given two equally sized arrays A and B of size N. A is empty and B has some values.
You need to fill A with an element X such that X belongs to B.
The only operations allowed are:
1. Copy B[i] to A[i].
2. Cyclic shift B by 1 to the the right.
You need to minimise the number of operations.
The first line contains a single positive integer N(1?≤?N?≤?106), denoting the size of the arrays.
Next line contains N space separated positive integers denoting the elements of the array B(1?≤?B[i]?≤?105).
Output a single integer, denoting the minimum number of operations required.
3
1 2 3
5
6
1 1 2 2 3 3
10
In the first test case:
We can have 5 steps as: fill first element, shift, fill second element, shift, fill third element.
Initially, A?=?[_,?_,?_],?B?=?[1,?2,?3]
After step 1, A?=?[1,?_,?_],?B?=?[1,?2,?3]
After step 2, A?=?[1,?_,?_],?B?=?[3,?1,?2]
After step 3, A?=?[1,?1,?_],?B?=?[3,?1,?2]
After step 4, A?=?[1,?1,?_],?B?=?[2,?3,?1]
After step 5, A?=?[1,?1,?1],?B?=?[2,?3,?1]
求每个数中相邻区间最大的最小的那个数。
1 #include <iostream> 2 #include <stdio.h> 3 #define ll long long 4 #define INF 0x3f3f3f3f 5 using namespace std; 6 const int N = 1e6+10; 7 int a[N], b[N/10+10][5]; 8 int main() { 9 int n; 10 cin >> n; 11 for(int i = 1; i <= n; i ++){ 12 scanf("%d", &a[i]); 13 if(!b[a[i]][0]) b[a[i]][1] = b[a[i]][2] = b[a[i]][3] = i; 14 else { 15 b[a[i]][2] = b[a[i]][3]; 16 b[a[i]][3] = i; 17 b[a[i]][4] = max(b[a[i]][4], b[a[i]][3] - b[a[i]][2] - 1); 18 } 19 b[a[i]][0] ++; 20 } 21 int ans = INF; 22 for(int i = 1; i <= n; i ++) { 23 b[a[i]][4] = max(b[a[i]][4], n - b[a[i]][3] + b[a[i]][1] - 1); 24 ans = min(ans, b[a[i]][4] + n); 25 } 26 printf("%d\n",ans); 27 return 0; 28 }
标签：mis stdio.h rate values integer 区间 minimum elements cin
原文地址：http://www.cnblogs.com/xingkongyihao/p/7616563.html
