HDU 2817 A sequence of numbers 等比OR等差-优快云博客

本文链接：https://blog.youkuaiyun.com/wust_ZJX/article/details/44314869

本文介绍了一种快速解决特定类型数列（等差或等比）中任意项的算法。该算法通过输入前三项和目标项的位置来计算数列中该项的值，并返回结果模200907后的值。

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A sequence of numbers

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3796 Accepted Submission(s): 1164

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

看了样例，YY一下，感觉要么等差要么等比，然后一写一交就过啦，(*^__^*) 嘻嘻……

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#define mod 200907
#define ll __int64
#define N 100
using namespace std;

ll fun(ll m,ll n)
{
    ll b=1;
    while(n)
    {
        if(n&1)
        b=(b*m)%mod;
       n=n>>1;
        m=(m*m)%mod;
    }
    return b;
}

int main()
{
    ll T;
    ll a,b,c,k;
    while(~scanf("%I64d",&T))
    {
        while(T--)
        {
            scanf("%I64d %I64d %I64d %I64d",&a,&b,&c,&k);

            ll d;
            ll ans;
            if(b-a==c-b)
            {
                d=b-a;
                ans=(d*k-d+a)%mod;
            }

            else if(b/a==c/b)
            {
                d=b/a;
                ans=(a*fun(d,k-1))%mod;
            }
             printf("%I64d\n",ans);

        }
    }
    return 0;
}