HDU 4908——BestCoder Sequence，hdu4908

HDU 4908——BestCoder Sequence，hdu4908

BestCoder Sequence Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 169    Accepted Submission(s): 82


Problem DescriptionMr Potato is a coder.
Mr Potato is the BestCoder.

One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as  Bestcoder Sequence.

As the best coder, Mr potato has strong curiosity, he wonder the number of consecutive sub-sequences which are  bestcoder sequences in a given permutation of 1 ~ N.
 
InputInput contains multiple test cases. 
For each test case, there is a pair of integers N and M in the first line, and an permutation of 1 ~ N in the second line.

[Technical Specification]
1. 1 <= N <= 40000
2. 1 <= M <= N
 
OutputFor each case, you should output the number of consecutive sub-sequences which are the  Bestcoder Sequences. 
 
Sample Input

1 115 34 5 3 2 1

 
Sample Output

13HintFor the second case, {3},{5,3,2},{4,5,3,2,1} are Bestcoder Sequence.

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#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define M 100001#define base 40000using namespace std;int a[M],dp[M];int main(){    int n,m,x;    while(scanf("%d%d",&n,&m)!=EOF){        for(int i=0;i<n;++i){            scanf("%d",&a[i]);            if(a[i]==m) x=i;        }        memset(dp,0,sizeof dp);        int ans=0;        for(int i=x+1;i<n;++i){            if(a[i]>m) ans++;  //大的加小的減            else ans--;            dp[ans+base]++;    //記錄出現該狀態的次數        }        ans=++dp[base];//當狀態數為base，才滿足中位元        int tmp=0;        for(int i=x-1;i>=0;--i){            if(a[i]>m) tmp++;            else tmp--;            ans+=dp[-tmp+base];//狀態相加為base的個數        }        cout<<ans<<endl;    }    return 0;}

1242

杭電ACM1005 Number Sequence

這個題跟智商有關係，是數學題。
有重複持續時間：49 我以前也是別人跟我說的
#include<stdio.h>
int main()
{
int f[1000]={0},n,i,a,b;
while(scanf("%d%d%d",&a,&b,&n)!=EOF)
{
f[0]=1;
f[1]=1;
if(a==0&&b==0&&n==0)
break;
for(i=2;i<(n%49);i++)
{
f[i]=(a*f[i-1]+b*f[i-2])%7;
}
printf("%d\n",f[n%49-1]);
}
return 0;
}
 
Max Sum http://acmhdueducn/showproblemphp?pid=1003 Problem Description Given a sequence a[1],a

上機調試去。