CF刷题-Codeforces Round #481-D. Almost Arithmetic Progression

题目链接：https://codeforces.com/contest/978/problem/D

题解：

　　题目的大意就是:这组序列能否组成等差数列？一旦构成等差数列，等差数列的公差必定确定，而且，对于给定的数列，公差的可能性是已知的。

　　我的解决思路是：对于给定的数列，最后一位数 - 第一位数 / （n -1） 必定是公差，而对于我们要判断的原数列，其最后一位 - 第一位数的值是可以-2,-1,0,+1,+2的，穷举可能性一切公差，找出变动最小的情况。代码如下(有点小复杂，没优化)：

　　

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<iostream>
#include<vector>
#include<string>
#include<cmath>
#include<set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N = +;
int ini[N];
int ans[N];
int tofopr;
set<int> jg;
int isok(int n,int newc){
    int c = ans[];
    int t = ;
    for(int i = ;i<=n - ;i++){
        c+= newc;
        if(abs(ini[i] - c) > ) return -;
        if(ini[i] != c){
            t++;
        }
    }
    return t;
}

int main(void){
    int n;
    jg.clear();
    scanf("%d",&n);
    if(n <= ){
        int b;
        for(int i = ;i<=n;i++){
            scanf("%d",&b);
        }
        printf("0\n");
    }
    else{
        int b;
        for(int i = ;i<=n;i++){
            scanf("%d",&b);
            ini[i] = b;
        }
        int c = ini[n] - ini[];
        int start,end;
        for(int i = c -;i<=c+;i++){
            tofopr = ;
            if(i%(n-) == ){
                int newc = i / (n - );
                if(i == c - ){
                    start = ini[] + ;
                    end = ini[n] - ;
                    if(end >= ){
                        ans[] = start;
                        ans[n] = end;
                        tofopr = isok(n,newc);
                        if(tofopr >= ){
                            tofopr += ;
                            jg.insert(tofopr);
                        }
                    }
                }
                if(i == c -){
                    start = ini[];
                    end = ini[n] - ;
                    if(end >= ){
                        ans[] = start;
                        ans[n] = end;
                        tofopr = isok(n,newc);
                        if(tofopr >= ){
                            tofopr += ;
                            jg.insert(tofopr);
                        }
                    }
                    start = ini[] + ;
                    end = ini[n];
                    ans[] = start;
                    ans[n] = end;
                    tofopr = isok(n,newc);
                    if(tofopr >= ){
                        tofopr += ;
                        jg.insert(tofopr);
                    }
                }
                if(i == c){
                    start = ini[];
                    end = ini[n];
                    ans[] = start;
                    ans[n] = end;
                    if(tofopr >= ){
                        tofopr = isok(n,newc);
                        jg.insert(tofopr);
                    }
                }
                if(i == c + ){
                    start = ini[] -;
                    end = ini[n];
                    if(start >= ){
                        ans[] = start;
                        ans[n] = end;
                        tofopr = isok(n,newc);
                        if(tofopr >= ){
                            tofopr += ;
                            jg.insert(tofopr);
                        }

                    }
                    start = ini[];
                    end = ini[n]+;
                    ans[] = start;
                    ans[n] = end;
                    tofopr = isok(n,newc);
                    if(tofopr >= ){
                        tofopr += ;
                        jg.insert(tofopr);
                    }
                }
                if(i == c + ){
                    start = ini[] - ;
                    end = ini[n] + ;
                    if(start >= ){
                        ans[] = start;
                        ans[n] = end;
                        tofopr = isok(n,newc);
                        if(tofopr >= ){
                            tofopr += ;
                            jg.insert(tofopr);
                        }
                    }
                }

            }

        }
        if(jg.empty()){
            printf("-1");
        }
        else
        {
            auto it = jg.begin();
            printf("%d",*it);
        }
    }
    return ;
}

　　而神犇的代码如下，思路是一致的，但是神犇找最小的方法很特殊！代码如下：

　

#include <bits/stdc++.h>

using namespace std;
const int maxn = 2e5 + ;
typedef long long ll;

ll n, num[maxn];

void init() {
    scanf("%lld", &n);
    for (int i = ; i < n; i++)
        scanf("%lld", &num[i]);
}

ll get_ans(ll tol) {
    ll cnt[],st;
    cnt[] = cnt[] = cnt[] = ;
    for(int i=;i<;i++) {
        if(i == ) st = num[]-;
        else if(i == ) st = num[];
        else st = num[] + ;

        for (ll j = ; j < n; j++) {
            if(abs(num[j]-st) > ) {
                cnt[i] = INT_MAX;
                break;
            }
            cnt[i] += abs(num[j] - st);
            st += tol;
        }
    }
    if(cnt[] == cnt[] && cnt[] == cnt[] && cnt[] == INT_MAX) //无法组成等差数列返回-1
        return -;
    return min(cnt[], min(cnt[], cnt[]));
}

int main() {
    init();
    ll Min = INT_MAX,tol;
    tol = num[]-num[];
    ll ans = get_ans(tol);
    if(ans >= )
        Min = min(Min, ans);
    for (int i = ; i <= ; i++) {
        ans = get_ans(tol-i);
        if (ans >= )
            Min = min(Min, ans);

        ans = get_ans(tol+i);
        if (ans >= )
            Min = min(Min, ans);
    }

    if (Min == INT_MAX)
        printf("-1");
    else
        printf("%lld", Min);

    return ;
}