BNUOJ 34982 Beautiful Garden 2014北京邀请赛B （有意思的枚举题）

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题意: 在坐标轴上有一些树，现在要重新排列这些树，使得相邻的树之间间距相等。求最少移动的树的数量。

分析: 至少有两棵没有移动枚举两棵树然后看中间有多少不用移动的注意判断重合的情况

代码:

////  Created by TaoSama on 2015-09-25//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, a[N];int cal(int l, int r) {    int d = a[r] - a[l];    if(d == 0) return  n - (r - l + 1); //左右重合    int ret = INF;    for(int k = 2; k < n; ++k) { //枚举k个间距        int cnt = 2;        for(int i = l + 1; i < r; ++i) { //不重合            if(a[i] != a[i - 1] && a[i] != a[r] && (a[i] - a[l]) * k % d == 0)                ++cnt;        }        ret = min(ret, n - cnt);    }    return ret;}int main() {#ifdef LOCAL//  freopen("in.txt", "r", stdin);//  freopen("out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    int t; scanf("%d", &t);    int kase = 0;    while(t--) {        scanf("%d", &n);        for(int i = 1; i <= n; ++i) scanf("%d", a + i);        printf("Case #%d: ", ++kase);        if(n <= 2) {            puts("0");            continue;        }        int ans = INF;        sort(a + 1, a + 1 + n);        for(int i = 1; i <= n; ++i)            for(int j = i + 1; j <= n; ++j)                ans = min(ans, cal(i, j));        printf("%d\n", ans);    }    return 0;}

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