二进制枚举+容斥原理+中国剩余定理
#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std; #define MAXN 20
typedef long long LL; int n;
int s[MAXN];
LL a[MAXN], m[MAXN]; //a是余数,m是除数 LL extented_gcd(LL a,LL b, LL &x, LL &y) //扩展欧几里得
{
if(a == 0 && b == 0) return -1;
if(b == 0){
x = 1; y = 0; return a;
}
LL d = extented_gcd(b, a%b, y, x);
y -= a/b*x;
return d;
} LL mult(LL a, LL k, LL m){ //快速乘法
LL res = 0;
while(k){
if(k & 1LL) res = (res + a) % m;
k >>= 1;
a = (a << 1) % m;
}
return res;
} //x = ai(mod mi) mi之间互素
//M = m1 * m2 * ... * mi
//Mi'*Mi = 1 (mod mi) Mi = M/mi
//x = sigma(ai * Mi * M') % M
LL CRT(LL l, LL r) //中国剩余定理
{
LL M = 1, ans = 0;
for (int i = 0; i <= n; ++i)
if(s[i]) M *= m[i];
for(int i = 0;i <= n;i++)
{
if(s[i]){
LL Mi = M/m[i];
LL x,y;
extented_gcd(Mi, m[i], x,y); //求Mi*x=1(mod m[i])
x = (x%m[i] + m[i]) % m[i];
ans = (ans+mult(a[i]*Mi % M, x, M)) % M; //计算最小解
}
}
return (r+M-ans)/M - (l-1+M-ans)/M; //计算[l,r]间解的个数
} int main()
{
//freopen("in", "r", stdin);
int T, cas = 0;
scanf("%d", &T);
while(T--){
LL l, r;
scanf("%d%lld%lld", &n, &l, &r);
memset(s, 0, sizeof(s));
m[n] = 7; a[n] = 0; s[n] = 1;
for(int i = 0;i < n;i++)
scanf("%lld%lld", &m[i], &a[i]);
LL ans = 0;
int all = 1 << n;
for(int i = 0;i < all;i++){ //二进制枚举同余方程
int t = i, k = 0;
for(int j = 0;j < n;j++){
s[j] = t & 1;
t >>= 1;
k += s[j]; //计算同余方程个数
}
k = k & 1 ? -1 : 1;//容斥定理-奇减偶加
ans += 1LL * k * CRT(l, r);
}
printf("Case #%d: %lld\n", ++cas, ans);
}
return 0;
}