hdu 4990 Reading comprehension 二分 + 快速幂

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 ",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

思路：若n为奇数， ans(n) = 2^(n-1) + 2^(n-3) +...2^0

　　　若n为偶数， ans(n) = 2^(n-1) + 2^(n-3) +...2^1

 

由于 2^1+2^2+2^3+2^4 = (2^1+2^2) + 2^2*(2^1+2^2)  故只需要算一半就好了， 那么就可以递归处理，注意，若为奇数项，把最后一项算出来加上去就好了

n为奇数时的递归边界是：2^0 = 1;

n为偶数时的递归边界是：2^1 = 2;

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <set>
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1|1
using namespace std;
typedef long long LL;
const int N = 100005;
LL n, m, nn, mm;
LL pow_mod(LL b)
{
    LL res = 1, a = 2;
    while(b) {
        if(b & 1) res = res * a % m;
        a = (a % m) * a % m;
        b >>= 1;
    }
    return res;
}
LL sum1(LL k)
{
    if(k == 1) return 1;
    LL tmp, now;
    if(k & 1) tmp = pow_mod(k - 1);
    else tmp = pow_mod(k);

    now = sum1(k >> 1) % m;
    LL res = (now % m + (now * tmp) % m) % m;
    if(k & 1) res = res + pow_mod(k * 2 - 2) % m;
    return res;

}
LL sum2(LL k)
{
    if(k == 1) return 2;
    LL tmp, now;
    if(k & 1) tmp = pow_mod(k - 1);
    else tmp = pow_mod(k);

    now = sum2(k >> 1) % m;
    LL res = (now % m + (now * tmp) % m) % m;
    if(k & 1) res = res + pow_mod(k * 2 - 1) % m;
    return res;
}
int main()
{
    while(~scanf("%lld%lld", &n, &m))
    {
        LL ans = 0;
        if(n & 1) {
            nn = (n + 1) >> 1;
            ans = sum1(nn);
            printf("%lld
", ans % m);
        }
        else {
            mm = (n) >> 1;
            ans = sum2(mm);
            printf("%lld
", ans % m);
        }
    }
}