2016多校联合第四场 HDU5768

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给你[l,r]区间，问有多少数，是7的倍数，并且mod任何p[i]≠a[i]

容斥+中国剩余定理。我先吐槽下刘汝佳CRT板子，简直垃圾，太相信板子，半小时的题强行卡我3小时。。

分析一下题，再看看数据范围，一眼容斥。设Pi,Pj是两个素数。那么mod他们分别等于ai,aj的数在[1,lcm(Pi,Pj)]区间只有一个，我们用n减去这个数在加上lcm就把偏移差消掉了，然后直接除以lcm就是这个区间中符合当前枚举的条件的数量了。然后容斥去重。注意，不能直接传入总值除以7，我们需要把（0,7）丢进CRT里跑，这样才能把某些应该消掉的数消掉，否则要出错

板子啊，一定不要太相信板子！！！。

////  Created by Running Photon//  Copyright (c) 2016 Running Photon. All rights reserved.//#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <sstream>#include <set>#include <vector>#include <stack>#define ALL(x) x.begin(), x.end()#define INS(x) inserter(x, x,begin())#define ll long long#define CLR(x) memset(x, 0, sizeof x)using namespace std;const int inf = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = (1 << 16) + 10;const int maxv = 1e3 + 10;const double eps = 1e-9;int tot;void exgcd(ll a, ll b, ll& d, ll& x, ll& y) {    if(!b) {d = a; x = 1; y = 0;}    else {exgcd(b, a%b, d, y, x); y -= x * (a / b);}}ll inv(ll a, ll n) { //mod n aµÄÄæ    ll d, x, y;    exgcd(a, n, d, x, y);    return d == 1 ? (x + n) % n : -1;}/*x = ai(mod mi)m1 * m2 * ... * mi = Mmi之间互素x = sigma(ai * Mi * Mi(-1)) % MMi = M / miMi(-1) = Mi 模mi的逆*/ll mul(ll a, ll b, ll M) {    ll ret = 0;    while(b) {        if(b & 1) ret = (ret + a) % M;        a = a * 2 % M;        b >>= 1;    }    return ret;}ll CRT(ll a[], ll m[], ll n) {    ll M = 1;    ll ans = 0;    for(int i = 1; i <= n; i++)        M *= m[i];    for(int i = 1; i <= n; i++) {        ll Mi = M / m[i];        ll x, y, d;        exgcd(Mi, m[i], d, x, y);        ans = (ans + mul(x, mul(a[i], Mi, M), M)) % M;    }    if(ans < 0) ans += M;    return ans;}ll p[22];ll a[22];ll A[22], m[22], all;ll calc(ll n) {    int state = 0;    ll res = 0;    for(state = 1; state < 1 << tot; state++) {        all = 0;        ll di = 7;        int op = __builtin_popcount(state);        A[++all] = 0;        m[all] = 7;        for(int i = 0; i < tot; i++) {            if(state >> i & 1) {                di *= p[i];                A[++all] = a[i];                m[all] = p[i];            }        }        ll kk = CRT(A, m, all);        ll tmp = (n - kk + di) / di;        if(op & 1) res += tmp;        else res -= tmp;    }    return n / 7 - res;}int main() {#ifdef LOCAL    freopen("C:\\Users\\Administrator\\Desktop\\in.txt", "r", stdin);    freopen("C:\\Users\\Administrator\\Desktop\\out.txt","w",stdout);#endif//  ios_base::sync_with_stdio(0);    int T;    scanf("%d", &T);    int cas = 0;    while(T--) {        ll n, x, y;        scanf("%lld%lld%lld", &n, &x, &y);        tot = n;        for(int i = 0; i < n; i++) {            scanf("%lld%lld", &p[i], &a[i]);        }        if(n == 0) printf("Case #%d: %lld\n", ++cas, y / 7 - (x - 1) / 7);        else printf("Case #%d: %lld\n", ++cas, calc(y) - calc((x - 1)));    }    return 0;}

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