技術標籤：二分

>Link

ybtoj防具佈置

>解題思路

設 s i s_{i} si​為前 i i i個位置的防具總數， s i s_i si​我們可以運用數學知識 O ( n ) O(n) O(n)求出來

因為“整個防線上有且僅有一個位置有破綻或根本沒有破綻”，所以當 s i s_i si​為偶數時，前 i i i個位置上一定沒有破綻，為奇數則說明破綻一定在前 i i i個位置中
考慮二分防線的位置，找到第一個為奇數的 s i s_i si​，
如果 s m i d s_{mid} smid​為奇數，則答案在前 m i d mid mid之中
如果 s m i d s_{mid}

時間複雜度大概為 O ( 31 N ) O(31N) O(31N)

>程式碼

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define N 200010
#define LL long long
using namespace std;

int T, n;
LL s[N], e[N], d[N], l, r, mid, ans;

LL asksum (LL
 
 x)
{
	LL sum = 0;
	for (int i = 1; i <= n; i++)
	  if (s[i] <= x) sum += (min (x, e[i]) - s[i]) / d[i] + 1;
	return sum;
}
void work ()
{
	scanf ("%d", &n);
	for (int i = 1; i <= n; i++)
	  scanf ("%lld%lld%lld", &s[i], &e[i], &d[i]);
	if (asksum ((LL)2147483647
 
) % 2 == 0) {printf ("There's no weakness.\n"); return;}
	l = ans = 1, r = (LL)2147483647;
	while (l < r)
	{
		mid = (l + r) / 2;
		if (asksum (mid) % 2 == 1) r = mid;
		else l = mid + 1;
	}
	printf ("%lld %lld\n", r, asksum (r) - asksum (r - 1));
}

int main()
{
	scanf ("%d", &T);
	while (T--) work ();
	return 0;
}