8.8

#include<cstdio>
#include<cstring>
#include<string>
#include<math.h>
#include<iostream>
#include<algorithm>
#include<iomanip>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<cstdlib>
#define eps 1e-9
const double pi = acos(-1.0);
#define ll long long
const ll N = 1e8;
using namespace std;
struct poly
{
    int e;
    double c;
}ans[10000],p[10000];
int main()
{
    int n, e, c, m = -1, cnt = 0;//m记录a中最高次e的指数，
    map<int, double>q;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> e >> c;
        q[e] = c;//将多项式a的指数和系数值都存入map中
        if (i == 0)m = e;//找到最高次指数
    }
    cin >> n;
    for (int i = 0; i < n; i++)cin >> p[i].e >> p[i].c;
    while (m >= p[0].e)//若a中的最高次比b中最高次大
    {
        double change = q[m] / p[0].c;//a的最高次除以b最高次的系数比
        double diff = m - p[0].e;//指数比，减完之后的值
        if (fabs(change) >= 0.05)
        {
            ans[cnt].e = diff;//将系数和指数都存入ans数组中
            ans[cnt++].c = change;
            for (int i = 0; i < n; i++)
            {
                q[p[i].e + diff] -= change*p[i].c;//change乘以b更新a中的变化
            }
        }
        else m--;//就执行判断下一项
        while (fabs(q[m]) < 0.05&&m >= p[0].e)m--;
    }
    cout << cnt;
    if (!cnt)cout << " 0 0.0";
    for (int i = 0; i < cnt; i++)printf(" %d %.1lf", ans[i].e, ans[i].c);
    cout << endl;
    cnt = 0;
    while (m >= 0)
    {
        if (fabs(q[m]) >= 0.05)
        {
            ans[cnt].e = m;
            ans[cnt++].c = q[m];
        }
        m--;
    }
    cout << cnt;
    if (!cnt)cout << " 0 0.0";
    for (int i = 0; i < cnt; i++)printf(" %d %.1lf", ans[i].e, ans[i].c);
    return 0;
}

#include<iostream>
#include<algorithm>
#include<string.h>//memset()函数的头文件
using namespace std;
typedef long long LL;
LL acv[100100],sum[100100];
int main()
{
    memset(acv,0,sizeof(acv));
    memset(sum,0,sizeof(sum));
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%lld",&acv[i]);
    }
    sort(acv+1,acv+1+n);
    sum[0]=0;
    for(int i=1;i<=n;i++){
        sum[i] = sum[i-1]+acv[i];//一个累加过程
    }
    int a1 = n/2;//前半部分
    int b1 = n-n/2;//后半部分
    LL diff1 = sum[n] - 2*sum[a1];//减去前半部分
    LL diff2 = sum[n] - 2*sum[b1];//减去后半部分
    if(diff1>diff2){//比较哪一部分大
        printf("Outgoing #: %d\n",b1);
        printf("Introverted #: %d\n",a1);
        printf("Diff = %lld\n",diff1);
    }
    else
    {
        printf("Outgoing #: %d\n",a1);
        printf("Introverted #: %d\n",a1);
        printf("Diff = %lld\n",diff2);
    }
}