HDU 5014 Number Sequence 贪心 2014 ACM/ICPC Asia Regional Xi'an Online

尽可能凑2^x-1

#include <cstdio>
#include <cstring>
const int N = 100005;
int a[N], p[N];
int init(int x) {
	int cnt = 0;
	while(x > 1) {
		x /= 2;
		cnt ++;
	}
	return cnt + 1;
}
int main() {
	int n;
	while(~scanf("%d", &n)){
		for(int i = 0; i <= n; i ++) {
			scanf("%d", &a[i]);
		}
		memset(p, 0, sizeof p);
		long long ans = 0;
		for(int i = n; i >= 0; i --) {
			if(p[i] != 0) continue;
			if(i == 0) continue;
			int t = 1<<(init(i));
			if(t == 2 * i) {
				p[i] = i-1;
				p[i-1] = i;
				ans += 2 * (i^(i-1));
			} else if(t == i + 1) {
				p[i] = 0;
				p[0] = i;
				ans += 2*i;
			} else {
				if(p[t-i] == 0) {
					p[i] = t - i - 1;
					p[t-i-1] = i;
					ans += 2 * (t-1);
				}
			}
		}
		printf("%I64d\n", ans);
		for(int i = 0; i <= n; i ++) {
			if(i != 0) printf(" ");
			printf("%d", p[a[i]]);
		}puts("");
	}
	return 0;
}

HDU 5014 Number Sequence 贪心 2014 ACM/ICPC Asia Regional Xi'an Online