【CodeForces 627A】XOR Equation(位运算)
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Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair(a, b)?
The first line of the input contains two integers s andx (2 ≤ s ≤ 1012,0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.
Print a single integer, the number of solutions to the given conditions. If no solutions exist, print0.
9 5
4
3 3
2
5 2
0
In the first sample, we have the following solutions: (2, 7),(3, 6), (6, 3), (7, 2).
In the second sample, the only solutions are (1, 2) and(2, 1).
题目大意:给出两个数 a,b 的和 s ,XOR x,求有多少个 a,b
思路:表示渣渣还不会位运算,看了好久题解才懂
For any two integers a and b, we have , where is the xor and a&b is the bitwise AND. This is because is non-carrying binary addition. Thus, we can find a&b = (s - x) / 2 (if this is not an integer, there are no solutions).
Now, for each bit, we have 4 cases: , and . If , then ai = bi, so we have one possibility:ai = bi = ai&bi. If , then we must have ai&bi = 0 (otherwise we print 0), and we have two choices: ai = 1 and bi = 0 or vice versa. Thus, we can return 2n, where n is the number of one-bits in x. (Remember to subtract 2 for the cases a = 0 or b = 0 if necessary.)
a + b = ( a ^ b ) + ( a & b ) * 2 有了这个关系,可以枚举 x 的每一位,若 x 的第 i 位是 1,则 a=0,b=1 或 a=1,b=0 有两种情况,此时 a & b=0 ,如果 a & b = 0,那么 ans<<1 ,如果 a & b = 1 则不成立,结果为 0 。若 x 的第 i 位是 0 ,则 a=1,b=1 ( a & b = 1) 或 a=0,b=0 ( a & b = 0 ) 因为 a & b的值是确定的,所以此时只有 1 种情况即 a 和 b 的第 i 位是固定的。首先 s - x 必须是非负的并且 必须是偶数,如果 s == x,会有解 (0,s) 和 (s,0),与题目的 a,b 非负矛盾,所以 减2,( 0 ^ a = a)
#include <bits/stdc++.h>typedef long long ll;using namespace std;int main(){ ll s,x; cin>>s>>x; ll n=s-x; if (n < 0 || n%2){ cout<<"0"<<endl; return 0; } ll m=n>>1,ans=1; while(x){ if(x&1 && m&1){ cout<<"0"<<endl; return 0; } if(x&1) ans<<=1; x>>=1; m>>=1; } if(!n) ans-=2; cout<<ans<<endl; return 0;}
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