摘要：直接写个RMQ就能过。#include#include#include#include#include#define Max(a,b) ((a)>(b)?(a):(b))#define Min(a,b) ((a)<(b)?(a):(b))#define Maxn 60010using namespace std;int maxnum[Maxn][20],minnum[Maxn][20],n,Log[Maxn+10];int Log2(int x){ int num=0; x/=2; while(x) { num++; x/=2; } ... 阅读全文