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"传送门" Solution 虽然不是点分治但用类似点分治的方法不断接近正确结果 Code cpp // luogu judger enable o2 include include include include include define F(i,a,b) for(register int i 阅读全文
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"传送门" Solution 裸区间DP Code cpp include include include include include include include define F(i,a,b) for(register int i=(a);i M; char str[]="!WING",c 阅读全文
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"传送门" Solution 根据prufer序列做的题，具体可以看 "这里" 还知道了一种避免高精除的方法quq Code 阅读全文
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"传送门" Solution 由于重量只有三种情况，那么想到用差分约束。 由于范围比较小，想到可以floyed求差分约束，暴力求天平另一边 Code cpp include include include include include define F(i,a,b) for(register in 阅读全文
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"传送门" Solution 复习下tire(￣▽￣)／ 裸的可持久化tire，我用树剖求了下LCA Code cpp include include include include include define F(i,a,b) for(register int i=(a);ilen) retur 阅读全文
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"传送门" 吐槽洛谷难度标签qwq Solution 显然是一道神奇的DP，由于总钱数不变，我们只需要枚举前两个人的钱数就可知第三个人的钱数 DP的时候先枚举只用前k个币种，然后枚举前两个人的钱数，然后枚举转移即可 Code cpp include include include include i 阅读全文
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"传送门" Solution 可以发现实际上是把n分为几个循环节，然后找循环节的$lcm$是这次的排数 而$lcm$必然是一些最高次幂的质数的成积，那么就dp求一下所有情况就好了 PS:注意并不是必须要等于n小于n都行，因为可以在后面补1而$lcm$不变 Code cpp include inclu 阅读全文