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。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a, x) memset(a, x, sizeof(a)) 15 #define rg register 16 typedef long long ll; 17 typedef double db; 18 const int INF = 0x3f3f3f3f; 19 const db eps = 1e-8; 20 const int maxn = 2e5 + 5; 21 inline ll read() 22 { 23 ll ans = 0; 24 char ch = getchar(), last = ' '; 25 while(!isdigit(ch)) {last = ch; ch = getchar();} 26 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 27 if(last == '-') ans = -ans; 28 return ans; 29 } 30 inline void write(ll x) 31 { 32 if(x < 0) x = -x, putchar('-'); 33 if(x >= 10) write(x / 10); 34 putchar(x % 10 + '0'); 35 } 36 37 int n; 38 struct Node 39 { 40 ll s, e, d; 41 }t[maxn]; 42 43 ll calc(ll x) 44 { 45 ll ret = 0; 46 for(int i = 1; i <= n; ++i) 47 { 48 if(t[i].s > x) continue; 49 ret += (min(x, t[i].e) - t[i].s) / t[i].d + 1; 50 } 51 return ret; 52 } 53 54 int main() 55 { 56 int T = read(); 57 while(T--) 58 { 59 n = read(); ll Max = 0; 60 for(int i = 1; i <= n; ++i) t[i].s = read(), t[i].e = read(), t[i].d = read(), Max = max(Max, t[i].e); 61 ll L = 0, R = Max; 62 if(!(calc(R) & 1)) {puts("Poor QIN Teng:( "); continue;} 63 while(L < R) 64 { 65 ll mid = (L + R) >> 1; 66 if(calc(mid) & 1) R = mid; 67 else L = mid + 1; 68 } 69 write(L), space, write(calc(L) - calc(L - 1)), enter; 70 } 71 return 0; 72 }