思路：
将x,y的公有素因子找出了，最多10来个。
然后枚举x的指数，再分别考虑每个素因子，可以得到y对应指数要到多少才会被算到。
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>using namespace std;
typedef long long ll;
const int maxn = 2e5 + 7;
const int mod = 998244353;int p[maxn],v[maxn],cnt;
int num1[20],num2[20],cnt1[20],cnt2[20];
int Power[20];
int a,b,c,d,x,y;
int id;void getprime() {for(int i = 2;i < maxn;i++) {if(v[i] == 0) {p[++cnt] = i;v[i] = i;}for(int j = 1;j <= cnt && 1ll * i * p[j] < maxn;j++) {v[i * p[j]] = p[j];if(i % p[j] == 0) break;}}
}ll gcd(ll n,ll m) {return m == 0 ? n : gcd(m,n % m);
}void init() {getprime();scanf("%d%d%d%d%d%d",&a,&b,&c,&d,&x,&y);ll num = gcd(x,y);for(int i = 1;i <= cnt;i++) {int amt1 = 0,amt2 = 0;while(x % p[i] == 0) {x /= p[i];amt1++;}while(y % p[i] == 0) {y /= p[i];amt2++;}if(num % p[i] == 0) {num1[++id] = p[i];cnt1[id] = amt1;num2[id] = p[i];cnt2[id] = amt2;}if(x == 1 || y == 1) break;}if(num % x == 0 && x != 1) {num1[++id] = x;cnt1[id] = 1;num2[id] = x;cnt2[id] = 1;}for(int i = 1;i <= id;i++) Power[i] = 0;
}int qpow(int x,int n) {int res = 1;while(n) {if(n & 1) res = 1ll * res * x % mod;x = 1ll * x * x % mod;n >>= 1;}return res;
}void solve() {ll ans = 1;for(int i = a;i <= b;i++) {for(int j = 1;j <= id;j++) {ll amt1 = 1ll * i * cnt1[j]; //x第j个素数的幂ll amt2 = (amt1 + cnt2[j] - 1) / cnt2[j]; //y至少要几次幂才能大于等于x对应的amt1 %= mod - 1;if(amt2 > d) {Power[j] = Power[j] + 1ll * (c + d) * (d - c + 1) / 2 % (mod - 1) * cnt2[j] % (mod - 1);Power[j] %= mod - 1;} else if(amt2 <= c) {Power[j] = Power[j] + amt1 % (mod - 1) * (d - c + 1) % (mod -1);Power[j] %= mod - 1;} else {Power[j] = Power[j] + i % (mod - 1) * cnt1[j] % (mod - 1) * (d - amt2 + 1) % (mod - 1);Power[j] %= mod - 1;Power[j] = Power[j] + (amt2 - 1 + c) * (amt2 - c) / 2 % (mod - 1) * cnt2[j] % (mod - 1);Power[j] %= mod - 1;}}}for(int i = 1;i <= id;i++) {ans = ans * qpow(num1[i],Power[i]) % mod;}printf("%lld\n",ans);
}int main() {init();solve();return 0;
}