hdu 3434 Sequence Adjustment

Sequence Adjustment

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 477    Accepted Submission(s): 150

Problem Description

Given a sequence consists of N integers. Each time you can choose a continuous subsequence and add 1 or minus 1 to the numbers in the subsequence .You task is to make all the numbers the same with 
the least tries. You should calculate the number of the least tries 
you needed and the number of different final sequences with the least tries.

 

Input

In the first line there is an integer T, indicates the number of test cases.(T<=30)
In each case, the first line contain one integer N(1<=N<=10^6), 
the second line contain N integers and each integer in the sequence is between [1,10^9].
There may be some blank lines between each case.

 

Output

For each test case , output “Case d: x y “ where d is the case number 
counted from one, x is the number of the least tries you need and y 
is the number of different final sequences with the least tries.

 

Sample Input

222 461 1 1 2 2 2

 

Sample Output

Case 1: 2 3Case 2: 1 2
Hint
In sample 1, we can add 1 twice at index 1 to get {4,4},or minus 1 twice at index 2 to get {2,2}, or we can add 1 once at index 1 and minus 1 once at index 2 to get {3,3}. So there are three different final sequences.
 

 

Author

wzc1989

 

Source

2010 ACM-ICPC Multi-University Training Contest（1）——Host by FZU

 

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#include<cstdio>#include<cstring>#include<algorithm>using namespace std;#define L 1000100int a[L],b[L],p[L];long long min(long long x,long long y){    if (x<y) return x;    return y;}int main(){    int T;    int cas=1;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++)  scanf("%d",&a[i]);        int len=1;//缩        b[len]=a[1];        for(int i=2;i<=n;i++)  if(a[i]!=a[i-1])  b[++len]=a[i];        p[1]=0;        for(int i=2;i<=len;i++)  p[i]=b[i]-b[i-1];        long long sum=0,ans=0;        for(int i=2;i<=len;i++)        {            if(p[i]*sum<0)  ans+=min(abs(sum),abs(p[i]));            sum+=p[i];        }        sum=abs(sum);        ans+=sum;//处理单向的需要付出的代价        printf("Case %d: %I64d %I64d\n",cas++,ans,sum+1);    }    return 0;}

//这种我现在还是无法证明#include <iostream>#include <cstdio>#include <cstring>using namespace std;int A[1000005];int main(){    int cas,r=1;    scanf("%d",&cas);    while(cas--)    {        int N;        scanf("%d",&N);        for(int i=0; i<N; i++)            scanf("%d",&A[i]);        long long a=0,b=0;        for(int i=1; i<N; i++)        {            if (A[i]-A[i-1]>0) a+=A[i]-A[i-1];            else b-=A[i]-A[i-1];        }        int Max = max(A[0],A[N-1]), Min = min(A[0],A[N-1]);        printf("Case %d: %I64d %d\n",r++,(a>b?a:b),Max-Min+1);    }}