此题可以找到规律f(n) = 1! * 2! *...*n!, 如果直接打表的话,由于n比较大(10000000),所以会超内存,这时候就要用到离线处理数据,就是先把数据存起来,到最后在暴力一遍求解就行了,代码如下
代码一(超内存):
1 #include <stdio.h> 2 3 const long long mod = 1000000007; 4 const int N = 10000007; 5 long long a[N]; 6 7 int main() 8 { 9 a[0] = a[1] = 1; 10 for (int i = 2; i < N; i++) 11 { 12 a[i] = a[i - 1] * i % mod; 13 } 14 15 for (int i = 2; i < N; i++) 16 { 17 a[i] = a[i - 1] * a[i] % mod; 18 } 19 int n; 20 while (~scanf("%d", &n)) 21 { 22 printf("%I64d\n", a[n]); 23 } 24 return 0; 25 }
代码二(AC)
1 #include <stdio.h> 2 #include <vector> 3 #include <map> 4 const long long mod = 1000000007; 5 using namespace std; 6 7 int main() 8 { 9 int n; 10 vector<int> a; 11 map<int, long long> vis; 12 while (scanf("%d", &n) != EOF) 13 { 14 a.push_back(n); 15 vis[n] = -1; 16 } 17 long long prep = 1, nowp, prea = 1, nowa; 18 for (int i = 1; i <= 10000000; i++)//离线算法 19 { 20 nowp = prep * i % mod;//相当于求阶乘 21 prep = nowp; 22 nowa = nowp * prea % mod;//求n个阶乘的乘积 23 prea = nowa; 24 if (vis.count(i)) 25 { 26 vis[i] = nowa; 27 } 28 } 29 for (int i = 0; i < a.size(); i++)//暴力一遍 30 { 31 32 printf("%I64d\n", vis[a[i]]); 33 } 34 return 0; 35 }
离线算法确实挺神奇的,要深刻理解才能做题
时间: 2024-10-24 19:21:16