我不理，我还要发老八秘制小代码

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=1e5+10;
LL a[N],n,m,p1,s2,s1,r,r1=0,r2=0,mi;
int main(){
	//freopen("  .in","r",stdin);
	//freopen("  .out","w",stdout);
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&a[i]);
	}
	scanf("%d%d%d%d",&m,&p1,&s1,&s2);
	a[p1]+=s1;
	for(int i=1;i<=n;i++){
		if(i<m){
			r1+=(m-i)*a[i];
		}else if(i>m){
			r2+=(i-m)*a[i];
		}
	}
	mi=abs(r1-r2);
	r=m;
	for(int i=1;i<=n;i++){
		if(i<m){
			if(abs(r1+(m-i)*s2-r2)<mi){
				mi=abs(r1+(m-i)*s2-r2);
				r=i;
			}
		}else if(i>m){
			if(abs(r2-r1+(i-m)*s2)<mi){
				mi=abs(r2-r1+(i-m)*s2);
				r=i;
			}
		}
	}
	printf("%d",r);
	//fclose(stdin);
	//fclose(stdout);
	return 0;
}

就是这样，AC过，求赞

#include<bits/stdc++.h>
using namespace std;
long long n,a[100001],m,p1,s1,s2,ans,sum1,sum2,ans1,ans2,pos;
int main(void){
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    cin>>m>>p1>>s1>>s2;
    a[p1]+=s1;
    for(int i=1;i<m;i++)sum1+=a[i]*(m-i);
    for(int i=m+1;i<=n;i++)sum2+=a[i]*(i-m);
    ans=1000000000000000000;
    for(int i=1;i<=n;i++){
        if(i<m)ans1=sum1+s2*(m-i),ans2=sum2;
        else if(i>m)ans1=sum1,ans2=sum2+s2*(i-m);
        else ans1=sum1,ans2=sum2;
        if(abs(ans1-ans2)<ans)ans=abs(ans1-ans2),pos=i;
    }
    cout<<pos;
}

#include <bits/stdc++.h>
using namespace std;
long long n;
long long s[100005];
long long m,p,s1,s2,p2;
long long minn=1e19;
long long sum1,sum2;
long long t1,t2;
struct T{
	long long ju;
	long long re;
	long long he;
}a[100005];//结构体是真好用 
inline long long bijiao(long long x,long long y){
	if (x>=y){
		return x-y;
	}
	else{
		return y-x;
	}
}//计算距离 
int main(){
	cin>>n;
	for (long long i=1;i<=n;i++){
		cin>>s[i];
	}
	cin>>m>>p>>s1>>s2;
	for (long long i=1;i<=n;i++){
		a[i].re=s[i];//人数 
		a[i].ju=max(m,i)-min(m,i);//距离 
		a[i].he=a[i].ju*a[i].re;//气势和 
	}
	a[p].he+=(s1*a[p].ju);
	for (long long i=1;i<=n;i++){
		if (i<m){
			sum1+=a[i].he;
		}
		else{
			sum2+=a[i].he;
		}//计算两方的气势和 
	}
	for (long long i=1;i<=n;i++){
		t1=sum1;
		t2=sum2;
		if (i<m){
			t1+=(m-i)*s2;
		}
		else if(i>m){
			t2+=(i-m)*s2;
		}//枚举每个点，如果是左t1加，如果是右t2加 
		long long tmp=bijiao(t1,t2);//调用 
		if (minn>tmp){
			minn=tmp;
			p2=i;
		}//判断，因为这里是下标，不能用min() 
	}
	cout<<p2;
	return 0;
}

代码有点长，既然都看到这，就让你们那仟仟玉手麻烦点个赞在走嘛

这个题应该很好做吧？我之前直播也讲过，显然枚举每一个点接受了新的兵营以后去重新更新一次势力值，然后求一下绝对这最小的即可，需要注意 放在$m$那里不属于任何势力值。

#include <bits/stdc++.h>
#define inf 1e18
#define ll long long
using namespace std;
const int N = 1e5 + 5;
ll n, a[N], m, p1, s1, s2, ans;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
    }
    cin >> m >> p1 >> s1 >> s2;
    a[p1] += s1;
    ll sum1 = 0, sum2 = 0;
    for (int i = 1; i < m; i++) sum1 += a[i] * (m - i);
    for (int i = m + 1; i <= n; i++) sum2 += a[i] * (i - m);
    ll ans1 = 0, ans2 = 0;
    ans = inf;
    int pos;
    for (int i = 1; i <= n; i++)
    {
        if (i < m)
        {
            ans1 = sum1 + s2 * (m - i);
            ans2 = sum2;
        }
        else if (i > m)
        {
            ans1 = sum1;
            ans2 = sum2 + s2 * (i - m);
        }
        else
        {
            ans1 = sum1;
            ans2 = sum2;
        }
        if (abs(ans1 - ans2) < ans)
        {
            ans = abs(ans1 - ans2);
            pos = i;
        }
    }
    cout << pos;
    return 0;
}