版权声明: https://blog.csdn.net/weixin_39792252/article/details/81948664
D money
题意:从1-n一次经过,可以buy或者sell a[i]这样的商品,问交易最少的次数能获得的最大的利润。
题解:找最长的1 2 2 3这样的区间,最大利润就是最大的间最小的,这样的交易次数最小。
代码:
/*
*2018 nowcoder second D money
*
*题解:找上升区间,注意1 2 2 3 是一段,特判一下;
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn = 1e5+7;
int n;
ll a[maxn], vis[maxn];
void solve(ll &mp, ll &mn)
{
for(int i = 2; i <= n; i++)
if(a[i]>a[i-1]) vis[i] = vis[i-1];
else if(a[i]==a[i-1]&&vis[i-1]!=(i-1)) vis[i] = vis[i-1];
for(int i = 1, p; i <= n; i = p+1)
{
p = i+1; while(vis[p] == vis[i]) p++; p--;
if(i != p) {
mp += a[p] - a[i]; mn++;
}
}
}
int main()
{
int t; scanf("%d", &t);
while(t--)
{
memset(a, 0, sizeof(a));
memset(vis, 0, sizeof(vis));
scanf("%d", &n);
for(int i = 1; i <= n; i++) {
scanf("%lld", &a[i]);
vis[i] = i;
}
ll mp = 0, mn = 0;
solve(mp, mn);
printf("%lld %lld\n", mp, 2ll*mn);
}
return 0;
}