BestCoder Round #13(前两题)

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

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 657    Accepted Submission(s): 369

Problem Description

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.

 

Input

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).

 

Output

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

 

Sample Input

216

 

Sample Output

9258

 

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代码：

#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 0x3f3f3f3fusing 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;}

第二题我竟然直接向想暴力，简直too young  too simple。。这题思路是先保存所有操作，然后最后询问的时候我们就逆推这些操作，最后得到要询问的数的下标，那么就引刃而解了，还有一个难点是怎么找出操作后的下标关系，估计这个应该是最难的。其实可以看出，比如1 2 3 4 5 6 7 8   ，那么经过fun1的变换后得到的序列是是1 3 5 7 2 4 6 8，那么可以看出后一半的下标在变换前是（当前位置-一半位置）*2，那么前一半的位置的原来坐标是（当前位置-1）*2+1,，那么逆序操作就好做多了，就是n-id-1，那么就简单了，相当于是离线的操作吧，又长见识了。。。

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

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 603    Accepted Submission(s): 102

Problem Description

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];
}

 

Input

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.

 

Output

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

 

Sample Input

13 5O 1O 2Q 1O 3Q 1

 

Sample Output

24

 

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代码：

#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 1000000007using 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;}