刁钻的顾客

发布时间: 2017年7月3日 10:23   时间限制: 3000ms   内存限制: 128M

2
2 4
1 1
2
3 2
3 1

Yes
No

对于这道题目，直接暴力枚举，时间复杂度是O(3^(20))是会爆掉的，而如果我们采用折半枚举地方法，时间复杂度只有O(3^20 * LOG(3^20))

详细解释bai's白书上有例子。直接附AC代码：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long LL;
LL a[22],b[22],mi[33];
LL sa[int(1e7)];
LL sb[int(1e7)];
int cnta;
int cntb;
int n;
LL pow(LL x,LL p){LL ans = 1;while(p){if(p & 1){ans *= x;}x *= x;p >>= 1;}return ans;
}
void sp(LL arr[],int& cnt,int l,int r){int n = r - l + 1;for(int S = 0;S < pow(3,n);S++){int tmpS = S;int ans = 0;for(int i = 0;i < n;i++){int w = tmpS%3;ans += (2*w+1)*(a[i + l] - b[i + l]);tmpS /= 3;}arr[cnt++] = ans;}
}
int main(){mi[0] = 1;for(int i = 1;i <= 30;i++){mi[i] = mi[i-1] * 2;}while(~scanf("%d",&n)){cnta = cntb = 0;for(int i = 0;i < n;i++){scanf("%lld",&a[i]);}for(int i = 0;i < n;i++){scanf("%lld",&b[i]);}sp(sa,cnta,0,n/2);sp(sb,cntb,n/2+1,n-1);sort(sb,sb + cntb);for(int i = 0 ;i < cnta;i++){for(int t = 1;t <= 30;t++){LL t_mi = mi[t];int f = 0;if((sa[i] == t_mi && cntb == 0 )|| *lower_bound(sb,sb+cntb,t_mi - sa[i]) == t_mi - sa[i]){f = 1;}if(f){puts("Yes");goto end;}}}puts("No");end:;}return 0;
}