Sorting the Tombstones
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice URAL 1252
Description
There is time to throw stones and there is time to sort stones…
An old desolate cemetery is a long dismal row of nameless tombstones There are N tombstones of various shapes. The weights of all the stones are different. People have decided to make the cemetery look more presentable, sorting the tombstone according to their weight. The local custom allows to transpose stones if there are exactly K other stones between them.
Input
The first input line contains an integer N (1 ≤ N ≤ 130000). Each of the next N lines contains an integer X, the weight of a stone in grams (1 ≤ X ≤ 130000).
Output
The output should contain the single integer — the maximal value of K (0 ≤ K < N), that makes possible the sorting of the stones according to their weights.
Sample Input
| input | output |
|---|---|
5 30 21 56 40 17 |
1 |
题目大意：给你n个数，让你求中间隔K个数可以交换两边的数时，让这n个数有序，问这个K最大是多少。 如样例：K = 1，即 56 和 17可以交换位置，21 和 40可以交换位置。
解题思路：我们可以设每个数的起始位置是idx，有序时应在的位置是dst。那么 dst = idx + K*x。 K是要求的值，x表示某个整数。那么要让所有的数都能交换到达有序的位置，那么，dst[i] = idx[i] + K[i] * x[i]。那么我们要求的K，就是所有的GCD(K[i]*x[i] , ans)。 还要注意的是，顺序包括两种，递增和递减，结果取两种的最大值。
#include<stdio.h>
#include<algorithm>
#include<bits/stdc++.h>
#include<string.h>
#include<bitset>
#include<math.h>
#include<iostream>
using namespace std;
const int maxn = 1e6;
struct Stone{
int wei,idx;
}stones[maxn];
int GCD(int a,int b){
return b == 0? a : GCD(b,a%b);
}
bool cmp1(Stone a,Stone b){
return a.wei < b.wei;
}
bool cmp2(Stone a,Stone b){
return a.wei > b.wei;
}
int main(){
int n;
while(scanf("%d",&n)!=EOF){
for(int i = 1; i <= n; i++){
scanf("%d",&stones[i].wei);
stones[i].idx = i;
}
sort(stones+1,stones+1+n,cmp1);
int nn = 0, ans = 0, gcd = 0;
for(int i = 1; i <= n; i++){
int tmp = abs(i - stones[i].idx);
if(tmp){
nn++; gcd = GCD(gcd,tmp);
}
}
if(nn == 0){
ans = n -1;
}
ans = max(ans,gcd-1);
sort(stones+1,stones+1+n,cmp2);
nn = 0, gcd = 0;
for(int i = 1; i <= n; i++){
int tmp = abs(i-stones[i].idx);
if(tmp){
nn++;
gcd = GCD(gcd,tmp);
}
}
if(nn == 0){
ans = n-1;
}
ans = max(ans,gcd-1);
printf("%d\n",ans);
}
return 0;
}
/*
5
30
21
56
40
17
*/