2017 NOIP2017 day 1

Posted 2021-06-08 Ariel_

tags:

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pts: ?

T1: 100

T2: 0/40/50

T3: 30

T1

[NOIP2017 提高组] 小凯的疑惑

数论，不定方程，同余，中国剩余定理，构造，找规律

solution:

大表找规律： a * b - a - b

还有个 long long

T2

[NOIP2017 提高组] 时间复杂度

模拟，字符串，栈

solution

多组数据跑完一组直接 return 0 了 /jk ~~(我是benb)~~

不会写模拟qwq

/*
work by:Ariel_
O(1) = 0;
O(n ^ 1) = 1;
O(n ^ 2) = 2;…… 
*/
#include <iostream>
#include <cstring>
#include <cstdio>
#include <stack>
#include <string>
using namespace std;

int read(){
    int x = 0,f = 1; char c = getchar();
    while(c < \'0\'||c > \'9\') {if(c == \'-\') f = -1; c = getchar();}
    while(c >= \'0\' && c <= \'9\') {x = x*10 + c - \'0\'; c = getchar();}
    return x*f;
}
int m, L, T;//T为它给的时间复杂度 
string tim, code[110];
int calc(int &p, string s){//计算一个代码中变量的两个边界 
  	int ans = 0, len = s.size();
	while(s[p] < \'0\'|| s[p] > \'9\' && p <= len){	
		if(s[p] == \'n\'){p++; return 1234567;}
		p++;
	}
    while(s[p] >= \'0\' && s[p] <= \'9\'){ 
	   ans *= 10, ans += s[p] - \'0\', p++;
	}
	return ans;	
}
stack<int> s, S;
stack <char> c;
stack <int> id; 
void subtask(){
   int T = 0, cnt = 0, flag = 1, opt = 1, vis[27], o = 0, fag[110], jl[110];
   memset(jl, 0, sizeof jl);
   memset(fag, 0, sizeof fag);//记录这行程序能不能进去 
   memset(vis, 0, sizeof vis);
   int len = tim.size(), p = 3;
   if(tim[2] == \'n\'){//计算出他给出的时间复杂度 
       while(tim[p] < \'0\'|| tim[p] > \'9\') p++;
       while(tim[p] >= \'0\' && tim[p] <= \'9\') T *= 10, T += tim[p] - \'0\', p++;	  
	}
	int maxx = -0x3f3f3f3f;
    for(int i = 1; i <= L; i++) {
    	p = 4;
        if(code[i][0] == \'F\'){
           S.push(i); id.push(i);
           if(vis[code[i][2] - \'a\' + 1]){opt = 0;break;}
		   c.push(code[i][2]);//证明这个变量已经用过了
		   vis[code[i][2] - \'a\' + 1] = 1;  
           int x = calc(p, code[i]), y = calc(p, code[i]);
		   if(y - x > 100 && o == 0){cnt++; jl[i] = 1;maxx = max(cnt, maxx);}
		   else if(x > y) fag[i] = 1, o++;//该行程序不能进去
   	       s.push(i);
		}
        else if(code[i][0] == \'E\'){
           vis[c.top() - \'a\' + 1] = 0;//结束了之后变量就可以用了
           if(fag[id.top()]){o--, id.pop();}
		   else id.pop();
		   c.pop(); 
           if(S.empty()){flag = 0; break;} 
           else  S.pop();
           maxx = max(cnt, maxx);
           int tp = s.top(); s.pop();
           if(jl[tp]) cnt--;
		}
	}
	if(!opt) printf("ERR\\n");
	else if(!flag)printf("ERR\\n");
    else if(maxx == T) printf("Yes\\n");
    else printf("No\\n");
}
int main(){
   m = read();
   while(m--) {
   	  L = read();
   	  getline(cin, tim);
   	  for (int i = 1; i <= L; i++) getline(cin, code[i]);
	  int cnt_F = 0, cnt_E = 0;
	  for (int i = 1; i <= L; i++) {
	  	 if(code[i][0] == \'F\') cnt_F++;
	  	 else if(code[i][0] == \'E\') cnt_E++;
	  }
	  if (cnt_E != cnt_F) {printf("ERR\\n");} 
	  else subtask();
   }
   puts("");
   return 0;
}