HDU 5446 Lucas + 中国剩余定理 + 快速乘法
HDU 5446
题目链接：
http://acm.hdu.edu.cn/showproblem.php?pid=5446
题意：
求，n，m的范围很大
思路：
Lucas + 中国剩余定理 + 快速乘法。
Lucas定理用来求当n和m很大时的C（n,m）%p（p是素数）。证明网上有，具体公式就是
Lucas(n,m,p) = C(n%p,m%p）*Lucas(n/p,m/p,p)
中国剩余定理用来求同余方程。对方程n的最小解。
设，，则
因为本题数据会爆long long，所以需要用快速乘法，具体见代码，类似快速幂。
源码：
include
include
include
include
include
include
include
include
using namespace std;
define LL long long
const int MAXN = 15;
LL prime[MAXN];
LL fac[100005], inv[100005];
LL lv[MAXN];
void exceed_gcd(LL a, LL b, LL &ans, LL &x, LL &y)
{
if(a < b){
exceed_gcd(b, a, ans, y, x);
return;
}
if(b == 0){
x = 1, y = 0;
ans = a;
return;
}
else{
exceed_gcd(b, a % b, ans, y, x);
y -= x * (a / b);
}
}
LL ppow(LL a, LL x, LL mod)
{
LL ans = 1;
while(x){
// printf(“mod = %I64d, a = %I64d, x = %I64d, ans = %I64d\n”, mod, a, x, ans);
if(x & 1)
ans = (ans * a) % mod;
a = (a * a) % mod;
x >>= 1;
}
return ans;
}
LL C(LL n, LL m, LL mod)
{
if(n < m || n < 0 || m < 0) return 0;
// printf(“n = %I64d, m = %I64d, fac[n] = %I64d, in[m] = %I64d, inv[n - m] = %I64d\n”, n, m, fac[n], inv[m], inv[n - m]);
return fac[n] * inv[m] % mod * inv[n - m] % mod;
}
LL lucas(LL n, LL m, LL mod)
{
// printf(“n = %I64d, m = %I64d\n”, n, m);
if(n < m) {/printf(“n=%I64d,m=%I64d\n”,n,m);/return 0;}
if(n == 0 || m == 0) return 1;
// printf(“n = %I64d, m = %I64d, C = %I64d, lucas = %I64d\n”, n, m, C(n % mod, m % mod, mod), lucas(n / mod, m / mod, mod));
return C(n % mod, m % mod, mod) * lucas(n / mod, m / mod, mod) % mod;
}
LL mul(LL a, LL b, LL mod)
{
a = (a % mod + mod) % mod;
b = (b % mod + mod) % mod;
LL ans = 0;
while(b){
if(b & 1)
ans = (ans + a) % mod;
b >>= 1;
a <<= 1;
a = a % mod;
}
return ans;
}
int main()
{
// freopen(“1006.in”, “r”, stdin);
LL n, m, cnt;
int t;
scanf(“%d”, &t);
while(t–){
scanf(“%I64d%I64d%I64d”, &n, &m, &cnt);
LL M = 1;
LL d, y;
LL ans = 0;
for(int i = 0 ; i < cnt ; i++){
scanf(“%I64d”, &prime[i]);
M *= prime[i];
fac[0] = 1;
for(int j = 1 ; j < prime[i] ; j++)
fac[j] = fac[j - 1] * j % prime[i];
inv[prime[i] - 1] = ppow(fac[prime[i] - 1], prime[i] - 2, prime[i]);
for(int j = prime[i] - 2 ; j >= 0 ; j–)
inv[j] = (inv[j + 1] * (j + 1)) % prime[i];
// for(int j = 0 ; j < prime[i] ; j++)
// printf(“fac[%d] = %I64d, inv[%d] = %I64d\n”, j, fac[j], j, inv[j]);
lv[i] = lucas(n, m, prime[i]);
}
for(int i = 0 ; i < cnt ; i++){
LL M1 = M / prime[i];
LL M2;
exceed_gcd(prime[i], M1, d, y, M2);
// printf(“i = %d, M1 = %I64d, M2 = %I64d, M = %I64d lv[%d] = %I64d\n”, i, M1, M2, M, i, lv[i]);
// M2 = (M2 % M + M) % M;
ans = (ans + mul(mul(lv[i], M1, M), M2, M));
}
ans = (ans % M + M) % M;
printf(“%I64d\n”, ans);
}
return 0;
}