Reading comprehension
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 188 Accepted Submission(s): 82
Problem Description
Read the program below carefully then answer the question.
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>
const int MAX=100000*2;
const int INF=1e9;
int main()
{
int n,m,ans,i;
while(scanf("%d%d",&n,&m)!=EOF)
{
ans=0;
for(i=1;i<=n;i++)
{
if(i&1)ans=(ans*2+1)%m;
else ans=ans*2%m;
}
printf("%d\n",ans);
}
return 0;
}
Input
Multi test cases，each line will contain two integers n and m. Process to end of file.
[Technical Specification]
1<=n, m <= 1000000000
[Technical Specification]
1<=n, m <= 1000000000
Output
For each case，output an integer，represents the output of above program.
Sample Input
1 10 3 100
Sample Output
1 5
当i是偶数时f[i]=2*f[i-1] 否则f[i]=2*f[i-1]+1 那么我们考虑，f[2*i]=4*f[2*i-2]+2 这样偶项就成为了一个独立的递推数列。 令b[0]=0,b[i]=4*b[i-1]+2 对于f 当i为偶数时，计算b[i/2]就可以了。 当i为奇数时，计算b[i/2]*2+1 计算b[i]可以建立矩阵递推最后右边乘上列向量[4 0
(0,2) ,上面那一项就是b[i]了。