标签：eric 技术 getchar 数组 examples 并且 lag pow rap
You are given an array ??1,??2,…,???? and an integer ??
.
You are asked to divide this array into ??
non-empty consecutive subarrays. Every element in the array should be included in exactly one subarray. Let ?? be the index of subarray the ??-th element belongs to. Subarrays are numbered from left to right and from 1 to ??
.
Let the cost of division be equal to ∑??=1??(????⋅??
. For example, if ??=[1,−2,−3,4,−5,6,−7] and we divide it into 3 subbarays in the following way: [1,−2,−3],[4,−5],[6,−7], then the cost of division is equal to 1⋅1−2⋅1−3⋅1+4⋅2−5⋅2+6⋅3−7⋅3=−9
.
Calculate the maximum cost you can obtain by dividing the array ??
into ??
non-empty consecutive subarrays.
The first line contains two integers ??
and ?? (1≤??≤??≤3⋅105
).
The second line contains ??
integers ??1,??2,…,???? (|????|≤106
).
Print the maximum cost you can obtain by dividing the array ??
into ??
nonempty consecutive subarrays.
5 2 -1 -2 5 -4 8
15
7 6 -3 0 -1 -2 -2 -4 -1
-45
4 1 3 -1 6 0
8
题解：将n个数的数组分成k个连续的子数组，并且第i个子数组的权值为i，则我们可以用后缀和。首先我们一定每个数都得至少取一次，则我么一定要取a[1],然后将后面的n-1个数排序，
因为题目要求获得答案最大，并且这后n-1个数我们任意取k-1个都能保证符合题意中的分法，则为了保证答案最大我们就要取比较大的前k-1个了～～
#include<bits/stdc++.h> #include<iostream> #include<stdio.h> #include<iomanip> #include<stack> #include<queue> #include<algorithm> #include<cstring> #include<map> #include<vector> #include<numeric> #include<iterator> #include<cmath> #define mem(a,x) memset(a,x,sizeof(a)); using namespace std; int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); } int lcm(int a, int b) { return a * b / gcd(a, b); } const int INF = 0x3f3f3f3f; typedef long long ll; const int mod=1000000007; typedef pair<int,int>Pi; typedef pair<ll, ll>Pii; map<int,int>mp; map<int, char *>mp1; map<char *, int>mp2; map<char, int>mp3; map<string,int>mp4; map<char,int>mp5; const int maxn = 300010; ll a[maxn]; int read(){ int flag=1; int sum=0; char c=getchar(); while(c<‘0‘||c>‘9‘){ if(c==‘-‘)flag=-1; c=getchar(); } while(c>=‘0‘&&c<=‘9‘){ sum=sum*10+c-‘0‘; c=getchar(); } return sum*flag; } ll Read(){ int flag=1; ll sum=0; char c=getchar(); while(c<‘0‘||c>‘9‘){ if(c==‘-‘)flag=-1; c=getchar(); } while(c>=‘0‘&&c<=‘9‘){ sum=sum*10+c-‘0‘; c=getchar(); } return sum*flag; } ll quickmul(ll a,ll b){ ll ans=0; while(b){ if(b&1){ ans=(ans+a)%mod; } a=(a+a)%mod; b>>=1; } return ans; } ll quickpow(ll a,ll b){ ll ans=1; while(b){ if(b&1) ans=(ans*a)%mod; a=(a*a)%mod; b>>=1; } return ans; } int main() { int n,k; cin>>n>>k; for(int i=1;i<=n;i++)cin>>a[i]; for(int i=n;i>=1;i--)a[i]+=a[i+1]; sort(a+2,a+1+n,greater<ll>()); ll ans=0; for(int i=1;i<=k;i++)ans+=a[i]; cout<<ans<<endl; return 0; }
标签：eric 技术 getchar 数组 examples 并且 lag pow rap
原文地址：https://www.cnblogs.com/cherish-lin/p/10989001.html
