Bitwise XOR of Arithmetic Progression
Write a program that, given three positive integers x, y and z (x, y, z < 232, x ≤ y), computes the bitwise exclusive disjunction (XOR) of the arithmetic progression x, x + z, x + 2z, …, x + kz, where k is the largest integer such that x + kz ≤ y.
题解
https://blog.csdn.net/PoPoQQQ/article/details/46853933
求异或和当然要逐位考虑。第 (i) 位的结果是：
[(sum_{i=0}^{lfloorfrac{y-x}{z}
floor} lfloorfrac{x+iz}{2^i}
floor)mod 2
]
这个就是经典的类欧问题了。
[sum_{x=0}^N lfloorfrac{ax+b}{c}
floor = Nlfloorfrac{aN+b}{c}
floor-sum_{x=0}^{lfloorfrac{aN+b}{c}
floor-1} lfloorfrac{cx+c-b-1}{a}
floor
]
时间复杂度 (O(log^2 n))。
int64 calc(int64 a,int64 b,int64 c,int64 n){
if(a==0 or (a*n+b)/c==0)
return (a*n+b)/c*(n+1);
if(a>=c)
return a/c*n*(n+1)/2+calc(a%c,b,c,n);
if(b>=c)
return b/c*(n+1)+calc(a,b%c,c,n);
return (a*n+b)/c*n-calc(c,c-b-1,a,(a*n+b)/c-1);
}
int main(){
for(int64 x,y,z;scanf("%lld%lld%lld",&x,&y,&z)!=EOF;){
int64 ans=0;
for(int i=0;i<32;++i)
ans|=(calc(z,x,1LL<<i,(y-x)/z)&1)<<i;
printf("%lld
",ans);
}
return 0;
}