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
Problem Description
As one of the most powerful brushes, zhx is required to give his juniors
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
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
Input
Multiply test cases(less than
1000
EOF
For each case, there are two integers
n
p
1≤n,p≤1018
1000
EOF
For each case, there are two integers
n
p
1≤n,p≤1018
Output
For each test case, output a single line indicating the answer.
Sample Input
2 233 3 5
Sample Output
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
Source
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;
}
版权声明：本文内容由互联网用户自发贡献，该文观点仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请联系我们举报，一经查实，本站将立刻删除。
发布者：全栈程序员-站长，转载请注明出处：https://javaforall.net/115513.html原文链接：https://javaforall.net
