题解 | #异或#

由异或的性质，两个数 $x,yx,y$ 异或和等于 $00$ 当且仅当 $x=yx=y$。

所以在区间 $x\in [a,b],y\in [c,d]x\backslash in[a,b],y\backslash in[c,d]$ 中选择两个相同的数的 $x,yx,y$ 的概率：

${\displaystyle \frac{\min (b,d)-\max (a,c)+1}{(b-a+1)\times (d-c+1)}}\backslash dfrac\{\backslash min(b,d)-\backslash max(a,c)+1\}\{(b-a+1)\backslash times(d-c+1)\}$（重合段的长度除以总个数）

#include<cstdio>
#define int long long
int init(){
	char c = getchar();
	int x = 0, f = 1;
	for (; c < '0' || c > '9'; c = getchar())
		if (c == '-') f = -1;
	for (; c >= '0' && c <= '9'; c = getchar())
		x = (x << 1) + (x << 3) + (c ^ 48);
	return x * f;
}
void print(int x){
	if (x < 0) x = -x, putchar('-');
	if (x > 9) print(x / 10);
	putchar(x % 10 + '0');
}
int gcd(int x, int y){
    return y ? gcd(y, x % y) : x;
}
int mn(int x, int y){
    return x < y ? x : y;
}
int mx(int x, int y){
    return x > y ? x : y;
}
signed main(){
    int a, b, c, d;
    while (scanf("%lld%lld%lld%lld", &a, &b, &c, &d) != EOF) {
        int x = mn(b, d) - mx(a, c) + 1;
        if (x < 1) puts("0/1");
        else {
            int y = (b - a + 1) * (d - c + 1);
            int g = gcd(x, y);
            x /= g, y /= g;
            print(x), putchar('/'), print(y), putchar('\n');
        }
    }
}