标签：acm 算法 c 杭电
A sequence of numbers
http://acm.hdu.edu.cn/showproblem.php?pid=2817
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4046 Accepted Submission(s): 1242
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
Sample Output
Source
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#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
const int MOD=200907;
__int64 fun1(__int64 q,__int64 n)// a1*q^(n-1)//等比数列
{
__int64 ans=1,base=q;
while(n)
{
if(n&1)
ans=ans*base%MOD;
base=base*base%MOD;
n>>=1;
}
return ans%MOD;
}
__int64 fun2(__int64 a1,__int64 d,__int64 n)//a1+(n-1)*d;
{
return a1%MOD+((n-1)%MOD*d%MOD)%MOD;
}
int jude(__int64 num1,__int64 num2,__int64 num3)
{
if(num1+num3==2*num2)
return 0;
else
return 1;
}
int main()
{
int n;
scanf("%d",&n);
while(n--)
{
__int64 a1,a2,a3,K;
scanf("%I64d%I64d%I64d%I64d",&a1,&a2,&a3,&K);
if(jude(a1,a2,a3))
{
__int64 q=a3/a2;
printf("%I64d\n",(a1%MOD*fun1(q,K-1)%MOD)%MOD);
}
else
{
__int64 d=a3-a2;
printf("%I64d\n",fun2(a1,d,K)%MOD);
}
}
return 0;
}hdoj 2817 A sequence of numbers（快速幂取模）
标签：acm 算法 c 杭电
原文地址：http://blog.csdn.net/lh__huahuan/article/details/44905125
