D. Bits Reverse 2018CCPC桂林站

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题目描述

Now given two integers x and y, you can reverse every consecutive three bits in arbitrary number’s binary form (any leading zero can be taken into account) using one coin. Reversing (1,2,3) means changing it into (3,2,1).
Could you please find a way that minimize number of coins so that x = y? If you can, just output the minimum coins you need to use.

输入

The first line of input file contains only one integer T (1≤T≤10000) indicating number of test cases.
Then there are T lines followed, with each line representing one test case.
For each case, there are two integers x, y (0≤x,y≤1018) described above.

输出

Please output T lines exactly.
For each line, output Case d: (d represents the order of the test case) first. Then output the answer in the same line. If there is no way for that, print -1 instead.

样例输入

复制样例数据

样例输出

Case 1: -1
Case 2: 1
Case 3: 2

提示

Sample 1: Considering following two binary string:
0: 0 ...0000
3: 0 ...0011
There is no way to achieve the goal.
Sample 2: Considering following two binary string:
3: 0 ...0011
6: 0 ...0110
You can reverse the lowest three digits in 3 so that 3 is changed into 6.
You just need to perform one reverse so that the minimum coin you need to use is 1.

【题意】

T组输入，每组输入包含两个数字x和y（long long int）。对x有一种操作，可以将 x的二进制任意相邻的三个位逆置。

问最少进行多少次操作，可以使得x等于y

【分析】

首先将x和y分别转为二进制并存在数组里。

任意相邻三个位逆置，其实不影响中间那一位，也就是隔位交换。所以奇数位和偶数位分别考虑即可。

对于奇数位，x和y的1的个数若不相等，则无解；否则，设x中1的位置列表ai，y中1的位置列表bi。则

偶数位同理，ans加在一起即为答案。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<string>
#include<set>
#pragma warning(disable:4996)
using namespace std;
const int maxn = 1005;
const int INF = 0x3f3f3f3f;
typedef long long ll;
ll x, y;
int a[maxn], b[maxn];
int v[maxn], u[maxn];
int main() {
	int cas = 1;
	int t;
	scanf("%d", &t);
	while (t--) {
		scanf("%lld%lld", &x, &y);
		printf("Case %d: ", cas++);
		for (int i = 0; i < 63; i++) {
			a[i] = (1ll << i) & x;
			b[i] = (1ll << i) & y;
		}
		int cnt1 = 0;
		int cnt2 = 0;
		for (int i = 0; i < 63; i += 2) {
			if (a[i]) {
				u[cnt1++] = i;
			}
			if (b[i]) {
				v[cnt2++] = i;
			}
		}
		if (cnt1 != cnt2) {
			printf("-1\n");
			continue;
		}
		ll ans = 0;
		for (int i = 0; i < cnt1; i++) {
			ans += abs(u[i] - v[i]) / 2;
		}
		 cnt1 = 0;
		 cnt2 = 0;
		for (int i = 1; i <= 63; i += 2) {
			if (a[i]) {
				u[cnt1++] = i;
			}
			if (b[i]) {
				v[cnt2++] = i;
			}
		}
		if (cnt1 != cnt2) {
			printf("-1\n");
			continue;
		}
		for (int i = 0; i < cnt1; i++) {
			ans += abs(u[i] - v[i]) / 2;
		}
		printf("%lld\n", ans);
	}
	return 0;
}