arc100C 【三分】【贪心】

题目链接：https://arc100.contest.atcoder.jp/tasks/arc100_a

题意：
求对于所有的b，
的最小值

题解：
贪心：
显然问题可以转化成p[i]=a[i]-i, 求abs(p[i]-b)之和
枚举b在第k个位置大于p，则答案即为
$p[k+1..n]-p[1..k]+(2\ast k-n)\ast a[k/k+1]$再分析一下2*k-n的正负即可
代码：

// by Balloons
#include 
#include 
#include 
#include 
#define mpr make_pair
#define debug() puts("okkkkkkkk")
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)

using namespace std;

typedef long long LL;
#define int LL

const int inf = 1e9, maxn=2e5+5;

int n,a[maxn],sum[maxn];

signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]), a[i] -= i;
	sort(a+1,a+n+1);
	for(int i=1;i<=n;i++)sum[i] = sum[i-1]+a[i];
	
	LL ans=1e18;
	for(int k=1;k<=n;k++){
		int oth = sum[k] - (sum[n]-sum[k]);oth *= -1;
		int id = 2*k-n;
		if(id > 0 || k == n)ans = min(ans,oth+id*a[k]);
		else ans = min(ans,oth+id*a[k+1]);
//		printf("%lld\n",oth);
	}
	printf("%lld\n",ans);

	return 0;
}

三分：
显然原式对b是一个单峰函数，用三分即可解决

// by Balloons
#include 
#include 
#include 
#include 
#define mpr make_pair
#define debug() puts("okkkkkkkk")
#define rep(i,a,b) for(int (i)=(a);(i)<=(b);(i)++)

using namespace std;

typedef long long LL;
#define int LL

const int inf = 1e9,maxn=2e5+5;

int n,a[maxn];

int solve(int x){
	int r=0;
	for(int i=1;i<=n;i++)r += abs(a[i]-x);
	return r;
}

signed main(){
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)scanf("%lld",&a[i]), a[i] -= i;
	int l=-1e9, r=1e9;
	while(l<r-1){
		int lmid = (l+r)>>1, rmid = (lmid+r) >> 1;
		if(solve(lmid) > solve(rmid))l=lmid;
		else r=rmid;
	}
//	printf("%d %d\n",l,r);
	int ans;
	if(solve(l)<solve(r))ans=l;
	else ans=r;
//	printf("? %d\n",ans);
	printf("%lld\n",solve(ans));

	return 0;
}