此计算器的分数计算公式基于 这里 @ 提供的公式。

本来只有单人的，要改成多人，又懒得重构，于是加了一大堆 per[pp].

#include<bits/stdc++.h>
#include<windows.h>
#include<conio.h>
using namespace std;
char c;
struct contest{
	double c, k;
	int year, score, lx;
	double res;
};
struct person{
	string name;
	double maxn[9], maxp[9], dqf[9], ans1, ans2, ans3, ans4;
	int last[9], qzm[9];
	vector<contest>v;
	person(){
		for(int i = 1; i <= 8; i++)maxn[i] = maxp[i] = dqf[i] = last[i] = qzm[i] = ans1 = ans2 = ans3 = ans4 = 0;
		v.clear();
	}
}per[10001];
bool cmp(contest a, contest b){return a.year < b.year;}
int main(){
	string name, s;
	int id, x, y, lx, pc = 0;
	cout << "新款OI积分计算器(4种模式)\n按1确认";
	while(1){
		c = _getch();
		if(c == '1')break;
	}
	system("cls");
	cout << "请输入每个人的名字,输入end完成\n";
	for(int i = 1; i <= 10000; i++){
		cout << "#" << i << " ";
		cin >> name;
		if(name == "end")break;
		per[i].name = name;
		pc = i;
	}
	system("cls");
	cout << "请按[人ID 比赛名 年份 分数]格式输入每次比赛\n比赛名仅限CSP-J,CSP-S,NOIP,WC,APIO,NOI-SC,NOI,IOI\n输入0 end 0 0完成\n";
	for(int i = 1; i <= pc; i++)cout << "#" << i << " " << per[i].name << "\n";
	while(1){
		cin >> id >> s >> x >> y;
		if(s == "end"){
			if(id == 0 && x == 0 && y == 0)break;
			else cout << "输入错误\n";
		}else if(id < 1 || id > pc)cout << "输入错误\n";
		else if(s == "CSP-J")per[id].v.push_back({0.75, 0.998, x, y, 1});
		else if(s == "CSP-S")per[id].v.push_back({0.8, 0.9975, x, y, 2});
		else if(s == "NOIP")per[id].v.push_back({0.85, 0.9975, x, y, 3});
		else if(s == "WC")per[id].v.push_back({0.88, 0.997, x, y, 4});
		else if(s == "APIO")per[id].v.push_back({0.88, 0.997, x, y, 5});
		else if(s == "NOI-SC")per[id].v.push_back({0.88, 0.9985, x, y, 6});
		else if(s == "NOI")per[id].v.push_back({1, 0.9965, x, y, 7});
		else if(s == "IOI")per[id].v.push_back({2, 0.997, x, y, 8});	
		else cout << "输入错误\n";
	}
	system("cls");
	for(int pp = 1; pp <= pc; pp++){
		sort(per[pp].v.begin(), per[pp].v.end(), cmp);
		for(int i = 0; i < per[pp].v.size(); i++){
			per[pp].v[i].res = per[pp].v[i].year * per[pp].v[i].c / pow(per[pp].v[i].k, per[pp].v[i].score);
			per[pp].ans1 += per[pp].v[i].res;
			lx = per[pp].v[i].lx;
			per[pp].maxn[lx] = max(per[pp].maxn[lx], per[pp].v[i].res);
			per[pp].maxp[lx] = per[pp].v[i].res;
			if(per[pp].last[lx] == 0)per[pp].last[lx] = per[pp].v[i].year;
			per[pp].dqf[lx] += per[pp].v[i].res * (1 << (per[pp].v[i].year - per[pp].last[lx]));
			per[pp].qzm[lx] = (1 << (per[pp].v[i].year - per[pp].last[lx] + 1)) - 1;
		}
		for(int i = 1; i <= 8; i++)per[pp].ans2 += per[pp].maxn[i];
		for(int i = 1; i <= 8; i++)per[pp].ans3 += per[pp].maxp[i];
		for(int i = 1; i <= 8; i++)per[pp].ans4 += per[pp].dqf[i] / max(per[pp].qzm[i], 1);
	}
	for(int pp = 1; pp <= pc; pp++){
		cout << "#" << pp << " " << per[pp].name << "\n";
		printf("总和模式:%.2lf\n最大模式:%.2lf\n最近模式:%.2lf\n带权模式:%.2lf\n\n", per[pp].ans1, per[pp].ans2, per[pp].ans3, per[pp].ans4);
	}
	return 0;
}