这一次又只出了一题，第二题没有分析好，竟然直接copy代码，不过长见识了。。

第一题给了一些限制条件，自己没有分析好，就去乱搞，结果各种不对，后来有读题才发现。。暴力乱搞。。

题目：

Beautiful Palindrome Number

A positive integer x can represent as (a1a2…akak…a2a1)10 or (a1a2…ak?1akak?1…a2a1)10 of a 10-based notational system, we always call x is a Palindrome Number. If it satisfies 0<a1<a2<…<ak≤9, we call x is a Beautiful Palindrome Number.
Now, we want to know how many Beautiful Palindrome Numbers are between 1 and 10N.

The first line in the input file is an integer T(1≤T≤7), indicating the number of test cases.
Then T lines follow, each line represent an integer N(0≤N≤6).

For each test case, output the number of Beautiful Palindrome Number.

2
1
6

9
258

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=1000000+10;
char str[maxn];

bool judge(int k)
{
   sprintf(str,"%d",k);
   int len=strlen(str);
   for(int i=0;i<len/2-1;i++)
      if(str[i+1]<=str[i])  return false;
   if(len%2) if(str[len/2]<=str[len/2-1])  return false;
   for(int i=0;i<len/2;i++)
    {
        if(str[i]!=str[len-i-1]||str[i]=='0'||str[len-i-1]=='0')
             return false;
    }
    return true;
}

int main()
{
    int t,n,ri,cnt;
    scanf("%d",&t);
    while(t--)
    {
        cnt=1;
        ri=1;
        scanf("%d",&n);
        if(n==0)  printf("1\n");
        else
        {
            for(int i=1;i<=n;i++)
                ri=ri*10;
            for(int i=2;i<=ri;i++)
            {
                if(judge(i))
                   {
                      // printf("%d ",i);
                       cnt++;
                   }
            }
            printf("%d\n",cnt);
        }
    }
    return 0;
}

还有不知道为嘛用int一直wa，longlong就过，不是取余了吗？？

题目：

Operation the Sequence

You have an array consisting of n integers: a1=1,a2=2,a3=3,…,an=n. Then give you m operators, you should process all the operators in order. Each operator is one of four types:
Type1: O 1 call fun1();
Type2: O 2 call fun2();
Type3: O 3 call fun3();
Type4: Q i query current value of a[i], this operator will have at most 50.
Global Variables: a[1…n],b[1…n];
fun1() {
    index=1;
  for(i=1; i<=n; i +=2) 
    b[index++]=a[i];
  for(i=2; i<=n; i +=2)
    b[index++]=a[i];
  for(i=1; i<=n; ++i)
    a[i]=b[i];
}
fun2() {
  L = 1;R = n;
  while(L<R) {
    Swap(a[L], a[R]); 
    ++L;--R;
  }
}
fun3() {
  for(i=1; i<=n; ++i) 
    a[i]=a[i]*a[i];
}

The first line in the input file is an integer T(1≤T≤20), indicating the number of test cases.
The first line of each test case contains two integer n(0<n≤100000), m(0<m≤100000).
Then m lines follow, each line represent an operator above.

For each test case, output the query values, the values may be so large, you just output the values mod 1000000007(1e9+7).

1
3 5
O 1
O 2
Q 1
O 3
Q 1

2
4

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<vector>
#include<cmath>
#include<string>
#include<queue>
#define eps 1e-9
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;

const int maxn=100000+10;
int t,n,m,cnt,b,op[maxn];
ll a[maxn];


ll solve(int cnt,int val)
{
    int half=(n+1)/2;
    ll mul=0;
    for(int i=cnt;i>=1;i--)
    {
        if(op[i]==1)
        {
            if(val>half)  val=(val-half)*2;
            else val=(val-1)*2+1;
        }
        else if(op[i]==2)
            val=n-val+1;
        else
            mul++;
    }
    ll ans=a[val];
    for(int i=1;i<=mul;i++)
         ans=ans*ans%mod;
	return ans;
}

int main()
{
    char s[2];
    scanf("%d",&t);
    while(t--)
    {
        cnt=0;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) a[i]=i;
        while(m--)
        {
            scanf("%s%d",s,&b);
            if(s[0]=='O')
               op[++cnt]=b;
            else
            {
                ll ans=solve(cnt,b);
                printf("%I64d\n",ans);
            }
        }
    }
    return 0;
}

BestCoder Round #13(前两题)