• HDU 5446 Unknown Treasure Lucas+中国剩余定理+按位乘


    HDU 5446 Unknown Treasure

    题意:求C(n, m) %(p[1] * p[2] ··· p[k])     0< n,m < 1018

    思路:这题基本上算是模版题了,Lucas定理求C(n,m),再用中国剩余定理合并模方程,因为LL相乘会越界,所以用到按位乘。

      1 #include <iostream>
      2 #include <cstdio>
      3 #include <fstream>
      4 #include <algorithm>
      5 #include <cmath>
      6 #include <deque>
      7 #include <vector>
      8 #include <queue>
      9 #include <string>
     10 #include <cstring>
     11 #include <map>
     12 #include <stack>
     13 #include <set>
     14 #define LL long long
     15 #define eps 1e-8
     16 #define INF 0x3f3f3f3f
     17 #define MAXN 10005
     18 #define MAXK 15
     19 using namespace std;
     20 LL p[MAXK], mod[MAXK];
     21 LL quick_mod(LL x, LL y, LL mod){
     22     if (y == 0) return 1;
     23     LL res = quick_mod(x, y >> 1, mod);
     24     res = res * res % mod;
     25     if (y & 1){
     26         res = res * x % mod;
     27     }
     28     return res;
     29 }
     30 LL comb(LL n, LL m, LL p){
     31     LL res = 1;
     32     for (int i = 1; i <= m; i++){
     33         LL x = (n + i - m) % p;
     34         LL y = i % p;
     35         res = res * (x * quick_mod(y, p - 2, p) % p) % p;
     36     }
     37     return res;
     38 }
     39 LL lucas(LL n, LL m, LL p){
     40     if (m == 0) return 1;
     41     return lucas(n / p, m / p, p) * comb(n % p, m % p, p) % p;
     42 }
     43 
     44 LL exgcd(LL  a, LL  b, LL & x, LL & y)
     45 {
     46     if (b == 0){
     47         x = 1;
     48         y = 0;
     49         return a;
     50     }
     51     else{
     52         LL temp = exgcd(b, a % b, x, y);
     53         LL t = y;
     54         y = x - y * (a / b);
     55         x = t;
     56         return temp;
     57     }
     58 }
     59 LL multi(LL a, LL b, LL m){
     60     LL res = 0;
     61     while (b){
     62         if (b & 1){
     63             res = (res + a) % m;
     64         }
     65         a = (a + a) % m;
     66         b >>= 1;
     67     }
     68     return (res + m) % m;
     69 }
     70 LL china(LL p[], LL mod[], LL m){
     71     LL M = 1;
     72     for (int i = 1; i <= m; i++){
     73         M *= p[i];
     74     }
     75     LL x, y, res = 0;
     76     LL d;
     77     for (int i = 1; i <= m; i++){
     78         LL s = M / p[i];
     79         d = exgcd(p[i], s, d, y);
     80         res = (res + multi(multi(y, s, M), mod[i], M)) % M;
     81     }
     82     return res;
     83 }
     84 int main()
     85 {
     86 #ifndef ONLINE_JUDGE
     87     freopen("in.txt", "r", stdin);
     88     //freopen("out.txt", "w", stdout);
     89 #endif // OPEN_FILE
     90     int T;
     91     scanf("%d", &T);
     92     LL n, m, k;
     93     while (T--){
     94         scanf("%I64d%I64d%I64d", &n, &m, &k);
     95         for (int i = 1; i <= k; i++){
     96             scanf("%I64d", &p[i]);
     97             mod[i] = lucas(n, m, p[i]);
     98         }
     99         LL ans = china(p, mod, k);
    100         printf("%I64d
    ", ans);
    101     }
    102 }
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  • 原文地址:https://www.cnblogs.com/macinchang/p/4814490.html
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