Almost AP
time limit per test
2.0 s
memory limit per test
512 MB
input
standard input
output
standard output
An array, is called an "Almost Arithmetic Progression" (Almost AP) if you can modify at most 3 elements in the array such that it becomes an arithmetic progression.
Recall the definition of the arithmetic progression: For an array a1, a2, ..., an to be an AP, ai + 1 - ai = c for all 1 ≤ i < n, in which c is a constant (c can be either positive, or negative, or 0).
You are asked to turn an Almost AP into an exact AP by modifying at most 3 numbers.
Input
The first line contains an integer n. (3 ≤ n ≤ 105)
The second line contains n space-separated integers: a1, a2, ..., an. ( - 109 ≤ ai ≤ 109)
Output
Output an arithmetic progression consisting of n integers b1, b2, ..., bn. Array b differs from array a by at most three numbers. You should also make sure that |bi| ≤ 1018.
The input guarantees there is at least one solution. If there are multiple of them, you can output any one.
Examples
input
5 1 2 4 4 5
output
1 2 3 4 5
input
4 1 3 5 7
output
1 3 5 7
input
4 2 3 3 3
output
4 3 2 1
kblack 在做数螃蟹的实验。按照「螃蟹爷爷的秘诀」一书上介绍的：由于东方神秘力量的影响，螃蟹的个数是一个完美的等差数列，例如：−2,−1,0,1,2,… 。是的，kblack 的螃蟹比较神奇，有的时候螃蟹的个数可能会是负的。
kblack 每天都去池塘记录螃蟹的个数。但是 kblack 自己给自己放了最多三天的假（也有可能两天，有可能一天，有可能不放），放假的时候螃蟹个数的记录就是 kblack 口胡的，有的时候会假得有点过分。
现给出 kblack n 天的记录，记录中至多有三个数是错误的。请纠正错误并输出正确的实验记录。
思维题，写来写去wa，总是跟正确思路差了那么一点点。数列中出现的差最多次数的一定是等差数列的公差，然后就是枚举每一个数应不应该换的问题了，最后跟原数组比较，若是换了3次之内，就是符合答案的解，输出即可。
代码实现：
/*
Look at the star
Look at the shine for U
*/
#include<bits/stdc++.h>
#define ll long long
#define PII pair<int,int>
#define sl(x) scanf("%lld",&x)
using namespace std;
const int N = 1e6+5;
const int mod = 1e9+7;
const int INF = 0x3f3f3f3f;
const double PI = acos(-1);
ll inv(ll b){if(b==1)return 1; return (mod-mod/b)*inv(mod%b)%mod;}
ll fpow(ll n,ll k){ll r=1;for(;k;k>>=1){if(k&1)r=r*n%mod;n=n*n%mod;}return r;}
ll s[N],ans[N],t[N];
map <ll,ll> mmp;
int main()
{
ll n,i,j,k;
sl(n);
for(i = 1;i <= n;i++) sl(s[i]),t[i] = s[i];
if(n <= 4)
{
for(i = 1;i <= n;i++)
printf("%lld ",s[1]);
puts("");
return 0;
}
ll dis = 0,maxx = -1e18;
for(i = 1;i < n;i++) mmp[s[i+1]-s[i]]++;
map <ll,ll> :: iterator it;
for(it = mmp.begin();it != mmp.end();it++)
{
if(maxx < it->second)
{
maxx = it->second;
dis = it->first;
}
}
// cout<<dis<<"***"<<endl;
ll l = 1;
for(i = 2;i <= n;i++)
{
ll sum = 0;
if(s[i]-s[i-1] == dis)
{
// cout<<s[i]<<endl;
for(j = i;j <= n;j++) s[j] = s[j-1]+dis;
for(j = i-1;j;j--) s[j] = s[j+1]-dis;
for(j = 1;j <= n;j++) if(s[j] != t[j]) sum++;
if(sum > 3)
{
for(j = 1;j <= n;j++) s[j] = t[j];
}
else
break;
}
}
for(i = 1;i <= n;i++)
{
printf("%lld ",s[i]);
}
puts("");
}