[BJWC2008]秦腾与教学评估

嘟嘟嘟

二分好题。

怎么二分呢？反正我是没想出来。

看了题解。

因为只有一个为奇数的点，所以对于一个位置x，求出区间[0, x]的教总和，如果为奇数，说明x取大了；否则x取小了（妙啊）。

虽然答案在int内，但是L + R可能会爆int，导致有几个点TLE了，所以还是都开long long 吧。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = 2e5 + ;
 inline ll read()
 {
     ll ans = ;
     char ch = getchar(), last = ' ';
     while(!isdigit(ch)) {last = ch; ch = getchar();}
     while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
     if(last == '-') ans = -ans;
     return ans;
 }
 inline void write(ll x)
 {
     if(x < ) x = -x, putchar('-');
     if(x >= ) write(x / );
     putchar(x %  + '');
 }

 int n;
 struct Node
 {
     ll s, e, d;
 }t[maxn];

 ll calc(ll x)
 {
     ll ret = ;
     for(int i = ; i <= n; ++i)
     {
         if(t[i].s > x) continue;
         ret += (min(x, t[i].e) - t[i].s) / t[i].d + ;
     }
     return ret;
 }

 int main()
 {
     int T = read();
     while(T--)
     {
         n = read(); ll Max = ;
         for(int i = ; i <= n; ++i) t[i].s = read(), t[i].e = read(), t[i].d = read(), Max = max(Max, t[i].e);
         ll L = , R = Max;
         if(!(calc(R) & )) {puts("Poor QIN Teng:( "); continue;}
         while(L < R)
         {
             ll mid = (L + R) >> ;
             if(calc(mid) & ) R = mid;
             else L = mid + ;
         }
         write(L), space, write(calc(L) - calc(L - )), enter;
     }
     return ;
 }