Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1270    Accepted Submission(s): 512

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>

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

Output

For each case，output an integer，represents the output of above program.

Sample Input

1 10
3 100

Sample Output

1
5

Source

BestCoder Round #8

这个题就是需要用log(n)解决上面的程序问题。

然后我们找奇数项的关系。

f[2k+1] = 2*f[2k] + 1 (k>=1)

f[2k] = 2*f[2k-1]

代入可得 f[2k+1] = 4*f[2k-1]+1 => bi = 4*bi-1+1

bi = 4*bi-1+1

bi-1 = 4*bi-2+1

..

b3 = 4*b2 + 1

b2 = 4*b1 +1

可得bk = 1+4+4^2+....+4^k-1

k与n之间的映射是 k = (n+1)/2 然后带入模板算就OK。偶数的话乘2.

#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include<vector>
typedef long long LL;

LL mod;
LL pow_mod(LL a,LL n){
    LL ans = 1;
    while(n){
        if(n&1) ans=ans*a%mod;
        a= a*a%mod;
        n>>=1;
    }
    return ans;
}
 LL cal(LL p,LL n){  ///这里是递归求解等比数列模板 1+p+p^2...+p^n
    if(n==0) return 1;
    if(n&1){///(1+p+p^2+....+p^(n/2))*(1+p^(n/2+1));
         return (1+pow_mod(p,n/2+1))*cal(p,n/2)%mod;
    }
    else { ///(1+p+p^2+....+p^(n/2-1))*(1+p^(n/2+1))+p^(n/2);
         return (pow_mod(p,n/2)+(1+pow_mod(p,n/2+1))*cal(p,n/2-1))%mod;
    }
}

int main()
{
    LL n;
    while(scanf("%lld%lld",&n,&mod)!=EOF)
    {
        if(n==1&&mod==1) {
            printf("0\n");
            continue;
        }
        LL k = (n+1)/2;
        LL ans = cal(4,k-1);
        if(n&1){
            printf("%lld\n",ans);
        }else {
            printf("%lld\n",ans*2%mod);
        }
    }
    return 0;
}