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: {//rednext.color = 3;next.time = g;break;}case 2: {//yellownext.color = 1;next.time = r;break;}case 3: {//greennext.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: {//redreturn now.time;}case 2: {//yellowreturn now.time + r;}case 3: {//greenreturn 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: {//rednext.color = 3;next.time = g;break;}case 2: {//yellownext.color = 1;next.time = r;break;}case 3: {//greennext.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: {//redreturn now.time;}case 2: {//yellowreturn now.time + r;}case 3: {//greenreturn 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: {//rednext.color = 3;next.time = g;break;}case 2: {//yellownext.color = 1;next.time = r;break;}case 3: {//greennext.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: {//redreturn now.time;}case 2: {//yellowreturn now.time + r;}case 3: {//greenreturn 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？