考虑推柿子
最开始的想法是如果两个(t)在(mod B)意义下相等,那么只需要比较一下((t+left lfloor frac{t}{B} floor ight)mod A)就好了
显然(t=t\% B+B imes lfloor frac{t}{B} floor)
于是第一维就是$t%B+(B+1) imes lfloor frac{t}{B} floor $
也就是说如果(t\%B)的是相等的,那么只要((B+1) imes lfloor frac{t}{B} floor)在(mod A)意义下是相等的,那么两个(t)就是本质相同的
考虑求一下后面那个东西的循环节,显然是(frac{lcm(B+1,A)}{B+1}=frac{A}{gcd(b+1,A)})
再考虑到(t\%B)的循环节是(B),所以整个的循环节就是(frac{AB}{gcd(B+1,A)}),也就是说(t)和(t+frac{AB}{gcd(B+1,A)})是本质相同的
由此我们把这个问题转化成了一个(mod frac{AB}{gcd(B+1,A)})意义的区间覆盖,我们只需要把这些区间放上去覆盖就好了,最后就是求一下被覆盖的总面积,这个是一个非常普及的贪心问题
非常丢人的写错了区间覆盖
代码
#include <bits/stdc++.h>
#define re register
#define max std::max
#define LL long long
inline LL read() {
LL x = 0;char c = getchar();
while (c < '0' || c > '9') c = getchar();
while (c >= '0' && c <= '9') x = (x << 3ll) + (x << 1ll) + c - 48, c = getchar();
return x;
}
int n, m;
LL A, B, L;
const int maxn = 1e6 + 5;
struct Seg {LL l, r;} a[maxn], b[maxn << 1];
LL gcd(LL a, LL b) { return !b ? a : gcd(b, a % b); }
inline void add(LL x, LL y) { b[++m].l = x, b[m].r = y; }
inline int cmp(Seg A, Seg B) { return A.l == B.l ? A.r > B.r : A.l < B.l; }
int main() {
n = read(), A = read(), B = read();
for (re int i = 1; i <= n; i++) a[i].l = read(), a[i].r = read();
for (re int i = 1; i <= n; i++) L = max(L, a[i].r);L++;
LL r = gcd(A, B + 1);
if (A / r <= L / B) L = A / r * B;
for (re int i = 1; i <= n; i++) {
LL x = a[i].r / L - a[i].l / L;
if (x >= 2) {add(0, L - 1);break;}
if (x == 1)
add(0, a[i].r % L), add(a[i].l % L, L - 1);
else
add(a[i].l % L, a[i].r % L);
}
std::sort(b + 1, b + m + 1, cmp);
int p = 1;LL ans = 0;
for (re int i = 1; i <= m; i = p) {
LL T = b[i].r;
while (b[p].l <= T && p <= m) T = max(T, b[p].r), p++;
ans += T - b[i].l + 1;
}
printf("%lld
", ans);
return 0;
}