hdu 4565 So Easy!(构造矩阵快速幂)

So Easy!

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 842 Accepted Submission(s): 228



Problem Description

　　A sequence Sn is defined as:



Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.

　　You, a top coder, say: So easy!



Input

　　There are several test cases, each test case in one line contains four positive integers: a, b, n, m. Where 0< a, m < 215, (a-1)2< b < a2, 0 < b, n < 231.The input will finish with the end of file.

Output

　　For each the case, output an integer Sn.

Sample Input

2 3 1 2013
2 3 2 2013
2 2 1 2013

Sample Output

4
14
4

Source

2013
ACM-ICPC长沙赛区全国邀请赛——题目重现

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zhoujiaqi2010

题解:构造矩阵快速幂

#include<stdio.h>
long long a,b,n,m,c[2];
struct point{
long long a[2][2];
}e,res;
void init()
{
e.a[1][1]=2*a%m;
e.a[0][1]=(b-a*a)%m;
res.a[0][1]=res.a[1][0]=e.a[0][0]=0;
res.a[0][0]=res.a[1][1]=e.a[1][0]=1;
}
struct point mult(struct point x,struct point y)
{
struct point temp;
int i,j,k;

for(i=0;i<2;i++)
{
for(j=0;j<2;j++)
{
for(temp.a[i][j]=k=0;k<2;k++)
temp.a[i][j]=(temp.a[i][j]+x.a[i][k]*y.a[k][j]%m)%m;
}
}

return temp;
};
void solve()
{
n--;
while(n)
{
if(n&1) res=mult(res,e);
e=mult(e,e);
n>>=1;
}
}
void output()
{
c[0]=(2*a)%m;
c[1]=(2*a*a%m+2*b%m)%m;
c[0]=((c[0]*res.a[0][0])%m+c[1]*res.a[1][0])%m;
printf("%lld\n",(c[0]+m)%m);
}
int main()
{
while(scanf("%lld%lld%lld%lld",&a,&b,&n,&m)>0)
{
init();
solve();
output();
}

return 0;
}