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70分运行超时

#include <iostream>
using namespace std;

//结构体，记录每个路口状态和所剩时间
struct status {
    int color;
    int time;
};
int r, y, g;
status st[100001];


//函数，得到下一个灯色,所剩时间为最大
status nextOf(status now)
{
    status next = { 0,0 };
    switch (now.color)
    {
    case 1: {//red
        next.color = 3;
        next.time = g;
        break;
    }
    case 2: {//yellow
        next.color = 1;
        next.time = r;
        break;
    }
    case 3: {//green
        next.color = 2;
        next.time = y;
        break;
    }
    default:break;
    }
    return next;

}

//函数：需要再此灯处耗费多少时间
int cost(status now)
{
    switch (now.color)
    {
    case 1: {//red
        return now.time;
    }
    case 2: {//yellow
        return now.time + r;
    }
    case 3: {//green
        return 0;
    }
    default:return now.time;
    }

}


//设计函数，参数为当前状态，所剩时间和经过时间，返回末状态和所剩时间
status fun(status now, long long ltime)
{
    status t = { 0,0 };
    if (now.color == 0)
    {
        t.color = now.color;
        t.time = now.time;
        return t;
    }
    while (now.time < ltime)
    {
        ltime -= now.time;
        now = nextOf(now);
    }
    t.color = now.color;
    t.time = now.time - ltime;

    /*if (now.time >= ltime)
    {
        t.color = now.color;
        t.time = now.time - ltime;
    }
    else
    {
        t = fun(nextOf(now), ltime - now.time);
    }*/
    return t;

}




int main()
{
    cin >> r >> y >> g;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> st[i].color >> st[i].time;
    }

    long long sum = 0;//经过时间
    for (int i = 0; i < n; i++)
    {
        if (st[i].color == 0)
        {
            sum += st[i].time;
            for (int j = i + 1; j < n; j++)
            {
                st[j] = fun(st[j], st[i].time);
            }
        }
        else
        {

            sum += cost(st[i]);
            for (int j = i + 1; j < n; j++)
            {
                st[j] = fun(st[j], cost(st[i]));
            }
        }
    }
    cout << sum << endl;;
    return 0;
}

满分瞅他人的，

#include <iostream>
using namespace std;
//结构体，记录每个路口状态和所剩时间
struct status {
    int color;
    int time;
};
int r, y, g;
status st[100001];
//函数，得到下一个灯色,所剩时间为最大
status nextOf(status now)
{
    status next = { 0,0 };
    switch (now.color)
    {
    case 1: {//red
        next.color = 3;
        next.time = g;
        break;
    }
    case 2: {//yellow
        next.color = 1;
        next.time = r;
        break;
    }
    case 3: {//green
        next.color = 2;
        next.time = y;
        break;
    }
    default:break;
    }
    return next;

}
//函数：需要再此灯处耗费多少时间
int cost(status now)
{
    switch (now.color)
    {
    case 1: {//red
        return now.time;
    }
    case 2: {//yellow
        return now.time + r;
    }
    case 3: {//green
        return 0;
    }
    default:return now.time;
    }

}
//设计函数，参数为当前状态，所剩时间和经过时间，返回末状态和所剩时间
status fun(status now, long long ltime)
{
    status t = { 0,0 };
    if (now.color == 0)
    {
        t.color = now.color;
        t.time = now.time;
        return t;
    }
    while (now.time < ltime)
    {
        ltime -= now.time;
        now = nextOf(now);
    }
    t.color = now.color;
    t.time = now.time - ltime;

    /*if (now.time >= ltime)
    {
        t.color = now.color;
        t.time = now.time - ltime;
    }
    else
    {
        t = fun(nextOf(now), ltime - now.time);
    }*/
    return t;

}




int main()
{
    cin >> r >> y >> g;
    int n;
    cin >> n;
    long long sum = 0;//经过时间
    for (int i = 0; i < n; i++)
    {
        cin >> st[i].color >> st[i].time;

        if (st[i].color == 0)
        {
            sum += st[i].time;
        }
        else
        {
            sum+= cost(fun(st[i], sum%(r+g+y)));
            
        }
    }
    
    cout << sum << endl;;
    return 0;
}

就一个关键，将sum%（r+g+y）。迭代或者递归次数限制到了常数级

还有一个就是fun函数对于0即经过一段道路的处理

上面是迭代
下面是递归

#include <iostream>
using namespace std;

//结构体，记录每个路口状态和所剩时间
struct status {
    int color;
    int time;
};

int r, y, g;
status st[100001];


//函数，得到下一个灯色,所剩时间为最大
status nextOf(status now)
{
    status next = { 0,0 };
    switch (now.color)
    {
    case 1: {//red
        next.color = 3;
        next.time = g;
        break;
    }
    case 2: {//yellow
        next.color = 1;
        next.time = r;
        break;
    }
    case 3: {//green
        next.color = 2;
        next.time = y;
        break;
    }
    default:break;
    }
    return next;

}

//函数：需要再此灯处耗费多少时间
int cost(status now)
{
    switch (now.color)
    {
    case 1: {//red
        return now.time;
    }
    case 2: {//yellow
        return now.time + r;
    }
    case 3: {//green
        return 0;
    }
    default:return now.time;
    }

}


//设计函数，参数为当前状态，所剩时间和经过时间，返回末状态和所剩时间
status fun(status now, long long ltime)
{
    status t = { 0,0 };
    /*if (now.color == 0)
    {
        t.color = now.color;
        t.time = now.time;
        return t;
    }
    while (now.time < ltime)
    {
        ltime -= now.time;
        now = nextOf(now);
    }
    t.color = now.color;
    t.time = now.time - ltime;
*/
    
    
    
    if (now.color == 0)
    {
        t.color = now.color;
        t.time = now.time;
        return t;
    }
    if (now.time >= ltime)
    {
        t.color = now.color;
        t.time = now.time - ltime;
    }
    else
    {
        t = fun(nextOf(now), ltime - now.time);
    }
    return t;

}




int main()
{
    cin >> r >> y >> g;
    int n;
    cin >> n;
    long long sum = 0;//经过时间
    for (int i = 0; i < n; i++)
    {
        cin >> st[i].color >> st[i].time;

        if (st[i].color == 0)
        {
            sum += st[i].time;
        }
        else
        {
            sum+= cost(fun(st[i], sum%(r+g+y)));
            
        }
    }
    
    cout << sum << endl;;
    return 0;
}

递归调用耗时少点，偶然还是必然？