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BestCoder Round #13(前兩題)

編輯：C++入門知識

BestCoder Round #13(前兩題)

這一次又只出了一題，第二題沒有分析好，竟然直接copy代碼，不過長見識了。。

第一題給了一些限制條件，自己沒有分析好，就去亂搞，結果各種不對，後來有讀題才發現。。暴力亂搞。。

題目：

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, 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

2
1
6

Sample Output

9
258

Statistic | Submit | Clarifications | Back
代碼：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#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

第二題我竟然直接向想暴力，簡直too young  too simple。。這題思路是先保存所有操作，然後最後詢問的時候我們就逆推這些操作，最後得到要詢問的數的下標，那麼就引刃而解了，還有一個難點是怎麼找出操作後的下標關系，估計這個應該是最難的。其實可以看出，比如1 2 3 4 5 6 7 8   ，那麼經過fun1的變換後得到的序列是是1 3 5 7 2 4 6 8，那麼可以看出後一半的下標在變換前是（當前位置-一半位置）*2，那麼前一半的位置的原來坐標是（當前位置-1）*2+1,，那麼逆序操作就好做多了，就是n-id-1，那麼就簡單了，相當於是離線的操作吧，又長見識了。。。
還有不知道為嘛用int一直wa，longlong就過，不是取余了嗎？？
題目：


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    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, m(0.
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

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


 

Sample Output

2
4


 

Statistic | Submit | Clarifications | Back



代碼：

#include
#include
#include
#include
#include
#include
#include
#include
#include
#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;
}