POJ 2313

Sequence

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 3885 Accepted: 1537

Description

Given a sequence with N integers A(1), A(2), ..., A(N), your task is to find out a sequence B(1), B(2), ..., B(N), such that 

V = (|A(1) – B(1)| + |A(2) – B(2)| + ... + |A(N) – B(N)|) + (|B(1) – B(2)| + |B(2) – B(3)| + ... +|B(N-1) – B(N)|)

is minimum.

Input

The first line in the input contains an integer N (1 <= N <= 100). Then follow N lines, the i-th of which contains an integer A(i) (-10000 <= A(i) <= 10000).

Output

The output only contains an integer, which is the minimum value of V.

Sample Input

3358

Sample Output

5

这道题是一个数学归纳的贪心题，

最主要的就是根据a[i],a[i+1],b[i-1]三者之间的关系来推出b[i]的数值即可；

#include<iostream>#include<cstdio>#include<cmath>using namespace std;int main(){    int a[101],b[101];    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    b[1]=a[1];    for(int i=2;i<n;i++)    {        if(a[i]>b[i-1]&&a[i]>a[i+1])            b[i]=max(a[i+1],b[i-1]);        else if(a[i]<b[i-1]&&a[i]<a[i+1])            b[i]=min(a[i+1],b[i-1]);        else            b[i]=a[i];    }    b[n]=a[n];    int sum=0;    for(int i=1;i<=n;i++)    {        sum=sum+fabs(a[i]-b[i]);    }    for(int i=2;i<=n;i++)    {        sum=sum+fabs(b[i-1]-b[i]);    }    printf("%d\n",sum);    return 0;}