HDU 5768 中国剩余定理

题目链接：Lucky7

题意：求在l和r范围内，满足能被7整除，而且不满足任意一组，x mod p[i] = a[i]的数的个数。

思路：容斥定理+中国剩余定理+快速乘法。 (奇+ 偶-)

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

#define LL long long
#define FOR(i, n) for (int i=0; i<n; ++i)
LL l, r;

int extend_gcd(LL a, LL b, LL &x, LL &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    } else {
        int r = extend_gcd(b, a%b, y, x);
        y -= x*(a/b);
        return r;
    }
}

LL qMul(LL a, LL b, LL mod) {
    a %= mod;
    LL ret = 0;
    while(b) {
        if (b & 1) ret = (ret + a) % mod;
        b >>= 1;
        a = (a + a) % mod;
    }
    return ret;
}

LL CRT(LL *a, LL *m, int n) {
    LL M = 1;
    for (int i=0; i<n; ++i) M *= m[i];

    LL x = 0;
    LL d, y;
    for (int i=0; i<n; ++i) {
        //LL d, y;
        LL tm = M/m[i];
        extend_gcd(m[i], tm, d, y);
        x = (x + qMul( qMul(d, tm, M), a[i], M)) % M;  // 参数传递有顺序
    }
    x = (x + M) % M;
    return (r+M-x)/M - (l-1+M-x)/M; // 直接返回l r区间有多少个解。
}

LL a[20], m[20];
LL chu[20], rema[20];

int main() {
    freopen("1.in.cpp", "r", stdin);
    int t;
    scanf("%d", &t);
    int cas = 0;

    while(t--) {
        int n;
        scanf("%d%lld%lld", &n, &l, &r);
        FOR(i, n) {
            scanf("%lld%lld", &chu[i], &rema[i]);
        }
        LL tot = (1<<n);
        LL ans = r/7 - (l-1)/7;

        for (int i=1; i<tot; ++i) {
            //cout << i << "++++
";
            int cnt = 0;
            FOR (j, n) {
                if (i&(1<<j)) {
                    m[cnt] = chu[j];
                    a[cnt] = rema[j];
                    cnt++;
                }
            }
            m[cnt] = 7, a[cnt] = 0;
            cnt++;
            LL tans = CRT(a, m, cnt);
            cnt = (cnt&1) ? 1 : -1;
            ans += tans * cnt;
        }
        //cout << "++++
";
        printf("Case #%d: %I64d
", ++cas, ans);
    }
    return 0;
}