大家好,又见面了,我是全栈君。
zhx’s contest
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 448 Accepted Submission(s): 147
n
zhx thinks the
ith
i
zhx defines a sequence
{ai}
i
1:
a1..ai
2:
ai..an
He wants you to tell him that how many permutations of problems are there if the sequence of the problems’ difficulty is beautiful.
zhx knows that the answer may be very huge, and you only need to tell him the answer module
p
1000
EOF
For each case, there are two integers
n
p
1≤n,p≤1018
2 233 3 5
2 1HintIn the first case, both sequence {1, 2} and {2, 1} are legal. In the second case, sequence {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, {3, 2, 1} are legal, so the answer is 6 mod 5 = 1
思路:由题意能够求出答案为(2^n-2)%p
可是n。p都是LL型的,高速幂的时候会爆LL,所以这里要用到高速乘法,高速乘法事实上和高速幂差点儿相同。就是把乘号改为加号
注意:当n为1时。要输出1,而当p为1时要输出0。
AC代码:
#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define LL long long using namespace std; LL n, p; LL multi(LL a, LL b) { //高速乘法。事实上和高速幂差点儿相同 LL ret = 0; while(b) { if(b & 1) ret = (ret + a) % p; a = (a + a) % p; b >>= 1; } return ret; } LL powmod(LL a, LL b) { //高速幂 LL ret = 1; while(b) { if(b & 1) ret = multi(ret, a) % p; a = multi(a, a) % p; b >>= 1; } return ret; } int main() { while(cin >> n >> p) { if(p == 1) { cout << 0 << endl; } else if(n == 1) { cout << 1 << endl; } else { LL ans = powmod(2, n) - 2; if(ans < 0) ans += p; cout << ans << endl; } } return 0; }
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