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;}