HDOJ 5363 Key Set 【快速冪】

HDOJ 5363 Key Set 【快速冪】

題目鏈接 http://acm.hdu.edu.cn/showproblem.php?pid=5363

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題目就是要求從1~n中所有數組成的各子集中，子集元素和爲偶數的個數
1、兩個奇數可以組成一個偶數
2、集合中元素不存在重複
Solution
Let a be the number of even integers and b be the number of old integers. The answer is:

(-1是指去掉空集的情況)
得出結果爲2^(n-1)-1
所以輸入後直接快速冪即可

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
int T, n;
const int mod = 1000000007;
typedef long long ll;

ll quickPower (int a, int b, int mod) { // a ^ b % mod
    ll ans = 1; // answer
    ll aBin = a; // a^1, a^2, a^4, a^8 ...
//    while(b > 0) {
//        if(b & 1) ans = ans * aBin % mod; // only calculate when the value is 1
//        b >>= 1; // right shift
//        aBin = aBin * aBin %mod; // square
//    }
    for( ; b; b >>= 1){
        if(b & 1) ans = ans * aBin % mod; // only calculate when the value is 1
        aBin = aBin * aBin %mod; // square
    }
    return ans;
}

int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        printf("%d\n", quickPower(2, n-1, mod)-1);
    }

    return 0;
}

標程

#include <bits/stdc++.h>
typedef long long LL;
using namespace std;

int main() {
  int T; scanf("%d", &T);
  for (int _ = 0; _ < T; ++ _) {
    int n; scanf("%d", &n); -- n;
    LL r(1), a(2), mod = 1e9 + 7;
    for (; n; n >>= 1) {
      if (n & 1) r = r * a % mod;
      a = a * a % mod;
    }
    r = (r + mod - 1) % mod;
    printf("%lld\n", r);
  }
  return 0;
}