牛客算法周周练1 - Maximize The Beautiful Value(前缀和)
链接:https://ac.nowcoder.com/acm/contest/5086/A
来源:牛客网
时间限制:C/C++ 2秒,其他语言4秒
空间限制:C/C++ 131072K,其他语言262144K
64bit IO Format: %lld
空间限制:C/C++ 131072K,其他语言262144K
64bit IO Format: %lld
题目描述
Today HH finds a non-decreasing sequence(a1,a2....an,ai≤ai+1), he thinks it's not beautiful so he wants to make it beautiful.
To make it, HH will choose exactly one number and move it forward at least k steps(i.e. you can move ai to aj if k≤i−j), and then he defines the beautiful value F(n) as ![](https://uploadfiles.nowcoder.com/images/20170907/301599_1504797235409_382728925B45DEF94535080A5B2B511A)
HH asks you to calculate max(F(n))
输入描述:
The first line contains an positive integer T(1≤T≤10), represents there are T test cases.
For each test case:
The first line contains two positive integers n,k(1≤n≤105,1≤k<n),the length of the sequence ,the least steps you need to move.
The second line contains n integers a1,a2…an(1≤ai≤108) - the sequence.
输出描述:
For each test case, you should output the max F(n).
输入
3 5 3 1 1 3 4 5 5 2 1 1 3 4 5 5 1 1 1 3 4 5
输出
46 50 53
说明
In the first example, you can move the fifth number 4 for 3 steps and make the sequence become [4,1,1,3,5], then the beautiful value is 4×1+1×2+1×3+3×4+5×5=46.
You can also move the fifth number to make it become [1,5,1,3,4], the beautiful value is also 46.
In the second example, you can move the fifth number 5 for 2 steps and make the sequence become [1,1,5,3,4]
In the third example, you can move the second number 1 for 1 steps and then the sequence is still [1,1,3,4,5]
备注:
scanf is commended。
因为给出的序列是单调不降的,所以给出的顺序F(n)一定是最大的。
所以只需要移动最小的距离使影响最小,即移动 k 个单位。
考虑移动每个数对总和产生的影响,找到最大值即可。
1 #include <bits/stdc++.h> 2 typedef long long LL; 3 #define pb push_back 4 const int INF = 0x3f3f3f3f; 5 const double eps = 1e-8; 6 const int mod = 1e9+7; 7 const int maxn = 1e5+10; 8 using namespace std; 9 10 LL a[maxn]; 11 LL sum[maxn]; 12 13 int main() 14 { 15 #ifdef DEBUG 16 freopen("sample.txt","r",stdin); //freopen("data.out", "w", stdout); 17 #endif 18 19 int T; 20 scanf("%d",&T); 21 while(T--) 22 { 23 int n,m; 24 scanf("%d %d",&n,&m); 25 LL all = 0; 26 for(int i=1;i<=n;i++) 27 { 28 scanf("%d",&a[i]); 29 sum[i] = sum[i-1]+a[i]; 30 all+=i*a[i]; 31 } 32 LL ans = 0; 33 for(int i=1+m;i<=n;i++) 34 { 35 LL t = all+(sum[i-1]-sum[i-1-m])-a[i]*m; 36 ans=max(ans,t); 37 } 38 printf("%lld\n",ans); 39 } 40 41 return 0; 42 }
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