A sequence of numbers
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4384 Accepted Submission(s):
1374
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.
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.
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.
number module (%) 200907.
Sample Input
2
1 2 3 5
1 2 4 5
Sample Output
5
16
题意：给你一个序列的前三位，判断是等差数列还是等比数列，然后求出这个数列的第k项并输出第k项对200907取模
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
#define mod 200907
using namespace std;
LL fun(LL a,LL b)
{
LL ans=1;
//a=a%mod;
while(b)
{
if(b&1)
ans=(a*ans)%mod;
b/=2;
a=(a*a)%mod;
}
return ans;
}
int main()
{
int t;
LL x,y,x1,y1;
LL a,b,c,k;
LL ans;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld%lld%lld",&a,&b,&c,&k);
if(2*b==a+c)//等差数列
printf("%lld\n",(a+(c-b)*(k-1))%mod);
else //等比数列
printf("%lld\n",(((fun((c/b),k-1))%mod)*(a%mod))%mod);
}
return 0;
}