HDU5696 区间的价值

传送门

一句话题意:求1~n长度区间最大值与最小值之乘积的最大值(保证数据随机)

Time cost: 45min

Solution:

比较基本的题 

首先统计答案可以针对每个点更新所有长度 这样可以单调

然后...然后就不会了

发现是数据随机 所以可以乱搞

对于每个点作为最大值 类似尺取法的方式更新最小值和答案

这样一般情况下复杂度随数据波动性变化 单调性越强越凉

Code:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<iostream>
#define B puts("GGGGGGG");
#define ms(a,b) memset(a,b,sizeof a)
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
#define inf 2147483647
using namespace std;
typedef long long ll;
ll read() {
    ll as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 110003;
//head
int n;
ll a[N],dp[N];
int main() {
    // freopen("in.in","r",stdin);
    while(~scanf("%d",&n)) {
        ms(dp,0);
        rep(i,1,n) a[i] = read();
        rep(i,1,n) {
            dp[1] = max(dp[1],a[i] * a[i]);
            int l = i,r = i;
            ll minn = a[i];
            while(1) {
                if(r-l+1 == n) break;
                if(l > 1 && (r == n || a[l-1] > a[r+1])) minn = min(minn,a[--l]);
                else minn = min(minn,a[++r]);
                if(a[l] > a[i] || a[r] > a[i]) break;
                dp[r-l+1] = max(dp[r-l+1],a[i] * minn);
            }
        }
        rep(i,1,n) printf("%lld
",dp[i]);
    }
    return 0;
}