HDOJ A sequence of numbers

最新推荐文章于 2020-08-04 20:56:30 发布

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本文探讨了一种用于在序列中快速查找特定位置元素的算法优化，通过避免使用二分法导致的时间限制错误，采用迭代求解策略成功解决了问题，并提供了两种不同实现方式的代码分析。

A sequence of numbers

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 4 Accepted Submission(s) : 0

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Problem Description

Xinlv wrote some sequences on the paper a long time ago, they might be arithmetic or geometric sequences. The numbers are not very clear now, and only the first three numbers of each sequence are recognizable. Xinlv wants to know some numbers in these sequences, and he needs your help.

Input

The first line contains an integer N, indicting that there are N sequences. Each of the following N lines contain four integers. The first three indicating the first three numbers of the sequence, and the last one is K, indicating that we want to know the K-th numbers of the sequence.

You can assume 0 < K <= 10^9, and the other three numbers are in the range [0, 2^63). All the numbers of the sequences are integers. And the sequences are non-decreasing.

Output

Output one line for each test case, that is, the K-th number module (%) 200907.

Sample Input

2
1 2 3 5
1 2 4 5

Sample Output

5
16

Source

2009 Multi-University Training Contest 1 - Host by TJU

水题也很纠结，一开始用二分法求TLE了，后面还是还是用二分法求就AC了。哥郁闷了……

TLE 的代码

#include<iostream>
using namespace std;
const int mod=200907;
int main(){
    int n;
    __int64 a,b,c,k;
    scanf("%d",&n);
    while(n--){
         scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k);
         if(c-b==b-a)
               printf("%I64d\n",(a%mod+((k-1)%mod)*((b-a)%mod))%mod );
         else{
              __int64 ans=a;
              __int64 q=(b/a)%mod;
              k--;
              __int64 temp=k;
              if(temp==1){
                  printf("%I64d\n",a%mod);
                  continue;
              }
              while(k>1){
                   q=(q*q)%mod;
                   k/=2;
              }
             ans=((ans%mod)*(q%mod))%mod;
             if(temp%2==1)
                 ans=ans*((b/a)%mod);
             printf("%I64d\n",ans);
         }
    }
    //system("pause");
    return 0;
}

AC的代码

#include<iostream>
using namespace std;
const int mod=200907;
__int64 powhaha(__int64 a,__int64 n){
     if(n==1) return a;
     __int64 x=powhaha(a,n/2); //要考虑n的奇偶性
     __int64 ans=(x*x)%mod;
     if(n%2==1)
         ans=(ans*a)%mod;
     return ans;
}
int main(){
    int n;
    __int64 a,b,c,k;
    scanf("%d",&n);
    while(n--){
         scanf("%I64d%I64d%I64d%I64d",&a,&b,&c,&k);
         if(c-b==b-a)
               printf("%I64d\n",(a%mod+((k-1)%mod)*((b-a)%mod))%mod );
         else{
              __int64 ans;
              __int64 q=(b/a)%mod;
              ans=( (a%mod) * (powhaha(q,k-1)%mod) ) %mod;
             printf("%I64d\n",ans);
         }
    }
    //system("pause");
    return 0;
}