题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5014
4 2 0 1 4 3
20 1 0 2 3 4
HDU坑爹爆long long,换了__int64过了。想法很简单,把两个数二进制的0和1尽量补全,优先满足大的数就可以了。不过要找到区间。
代码:
#include <iostream> #include <cstdio> using namespace std; __int64 n, a[100010]; struct right { __int64 s, r, l; }rt[1000]; __int64 getNear(__int64 x) { __int64 z = 1; while(x) { x >>= 1; z <<= 1; } return z-1; } int main() { while(~scanf("%I64d", &n)) { __int64 m = n; rt[0].r = m; rt[0].s = getNear(m); rt[0].l = rt[0].s-rt[0].r; //cout << rt[0].l << " " << rt[0].r << " " << rt[0].s << endl; __int64 cnt = 0; while(1) { m = rt[cnt].l-1; if(m < 0) break; cnt++; rt[cnt].r = m; rt[cnt].s = getNear(m); rt[cnt].l = rt[cnt].s-rt[cnt].r; //cout << rt[cnt].l << " " << rt[cnt].r << " " << rt[cnt].s << endl; } for(__int64 i = 0; i <= n; i++) scanf("%I64d", &a[i]); //a[i] = i; __int64 t = 0; for(__int64 i = 0; i <= n; i++) for(__int64 j = 0; j <= cnt; j++) { if(a[i] >= rt[j].l && a[i] <= rt[j].r) { //cout << rt[j].l << " " << rt[j].r << " " << rt[j].s << endl; //printf("%d ", rt[j].s-a[i]); a[i] = rt[j].s-a[i]; t += rt[j].s; break; } } printf("%I64d\n", t); for(__int64 i = 0; i < n; i++) printf("%I64d ", a[i]); printf("%I64d\n", a[n]); } return 0; }
HDU 5014 Number Sequence(2014 ACM/ICPC Asia Regional Xi'an Online) 题解
原文地址:http://blog.csdn.net/u011439796/article/details/39274335