swjtu 2385 Maximize The Beautiful Value

Maximize The Beautiful Value

发布时间: 2017年5月14日 19:26   最后更新: 2017年5月14日 19:30   时间限制: 2000ms   内存限制: 128M

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 ∑ni=1i×ai.

HH asks you to calculate max(F(n))

输入

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

For each test case, you should output the max F(n).

样例输入1 复制

35 31 1 3 4 55 21 1 3 4 55 11 1 3 4 5

样例输出1

465053

提示

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 second example, you can move the  second number 1 for 1 steps and then the sequence is still [1,1,3,4,5]

scanf is commended。

题意：输入n,k，总共n个数，并且这n个数是非减的；选择一个数字最少要向前移动k个数，然后求出移动后的值；问移动哪一个数能够使得值为最大；求值：i*a[i] (i:1~n)；

思路：因为这n个数是非减的，列出几个样例就可以知道只向前移动k步是最优的，不能再向前多移动，否则值会减小；开辟三个数组：a[]:本来的数，b[]:对a[]的前缀和；c[i]:i*a[i]；S[i]:对c[]的前缀和；注意：处理的前缀和b[0]=0,S[0]=0;将数列分为3部分，比如n== 9,k==2;我们将第6个数向前移动k位应该是这样：

代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100005;typedef long long ll;ll a[maxn],b[maxn],c[maxn];ll S[maxn];int main(){    int T;    scanf("%d",&T);    while(T--)    {        memset(a,0,sizeof(a));        memset(b,0,sizeof(b));        memset(S,0,sizeof(S));        int n,k;        scanf("%d%d",&n,&k);        for(int i=1;i<=n;i++)        {            scanf("%lld",&a[i]);            b[i]=b[i-1]+a[i];            c[i]=i*a[i];            S[i]=S[i-1]+c[i];        }//        for(int i=1;i<=n;i++)//            printf("%lld ",b[i]);//        printf("\n");//        for(int i=1;i<=n;i++)//            printf("%lld ",c[i]);//        printf("\n");//        for(int i=1;i<=n;i++)//            printf("%lld ",S[i]);//        printf("\n");        ll ans=0,sum1=0,sum2=0,sum3=0,sum=0;        for(int i=k+1;i<=n;i++)        {            sum1=S[i-k-1];            sum2=S[i-1]-S[i-k-1]+b[i-1]-b[i-k-1]+c[i]-k*a[i];            sum3=S[n]-S[i];//printf("%lld %lld %lld \n",sum1,sum2,sum3);            sum=sum1+sum2+sum3;//            printf("%lld\n",sum);            ans=max(sum,ans);        }        printf("%lld\n",ans);    }    return 0;}