GCD XOR UVA

Think:
1埃式筛法思想建立约数表
2初始打表寻找运算的数学规律
3xor运算(不带进位的二进制加法)
运算法则

1. a ⊕ a = 02. a ⊕ b = b ⊕ a3. a ⊕b ⊕ c = a ⊕ (b ⊕ c) = (a ⊕ b) ⊕ c;4. d = a ⊕ b ⊕ c 可以推出 a = d ⊕ b ⊕ c.((a^b) == c <=> a(a^c) == b)5. a ⊕ b ⊕ a = b.6.若x是二进制数0101，y是二进制数1011；则x⊕y=1110只有在两个比较的位不同时其结果是1，否则结果为0即“两个输入相同时为0，不同则为1”！

xor运算——百度百科链接

vjudge题目链接

以下为Accepted代码

/*1打表发现数学规律gcd(a, b) == (a^b) => (a-b) == (a^b)2筛法思想建立约数表3xor运算:(a^b) = c <=> (a^c) = b;*/#include <cstdio>#include <cstring>using namespace std;int v[30000004] = {0};int main(){    int a, b, c;    ///筛法思想建立约数表    for(c = 1; c <= 15000000; c++){        for(a = c + c; a <= 30000000; a += c){            b = a - c;            if(c == (a^b))                v[a]++;        }    }    ///now:v[a]:(a, xi);    for(int i = 2; i <= 30000000; i++){        v[i] += v[i-1];    }    ///now:v[a]:(ai, xi);    int T, k, n;    scanf("%d", &T);    for(k = 1; k <= T; k++){        scanf("%d", &n);        printf("Case %d: %d\n", k, v[n]);    }    return 0;}