5 条题解
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1
#include<iostream> #include<string.h> using namespace std; const int INF=0x3f3f3f3f; const int N=2e5+10; long long n,x,a[N],len; int main(){ cin>>n; a[++len]=1; for(int i=1;i<=n;i++){ for(int j=1;j<=len;j++){ a[j]=a[j]*i+x; x=a[j]/10; a[j]%=10; } while(x){ a[++len]=x%10; x/=10; } } for(int i=len;i>=1;i--){ cout<<a[i]; } }
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0
#include <iostream> #include <vector> using namespace std; // 计算阶乘的函数,返回存储阶乘结果的数组,低位在前 vector<int> factorial(int n) { vector<int> result(1, 1); // 初始化为1,低位在前,方便计算 for (int i = 2; i <= n; ++i) { int carry = 0; for (size_t j = 0; j < result.size(); ++j) { int product = result[j] * i + carry; result[j] = product % 10; carry = product / 10; } while (carry > 0) { result.push_back(carry % 10); carry /= 10; } } return result; } int main() { int n; cin >> n; if (n == 0 || n == 1) { cout << 1 << endl; return 0; } vector<int> res = factorial(n); // 从高位到低位输出结果 for (auto it = res.rbegin(); it != res.rend(); ++it) { cout << *it; } cout << endl; return 0; }
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0
100%AC
不AC我倒立洗头#include<iostream> #include<cstring> #include<cmath> #include<iomanip> #include<queue> #include<algorithm> #include<vector> #include<stack> using namespace std; #define LL long long const int N=1e5+10; using namespace std; int n , ans[N] , len , x; int main() { cin>>n; ans[++len] = 1; for(int i = 2; i <= n; i++) { for(int j = 1; j <= len; j++) { ans[j] = ans[j] * i + x; x = ans[j] / 10; ans[j] %= 10; } while( x ) { ans[++len] = x % 10; x /= 10; } } for(int i = len; i >= 1; i --) cout << ans[i]; return 0; }
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#include<iostream> #include<string.h> using namespace std; const int INF=0x3f3f3f3f; const int N=2e5+10; long long n,x,a[N],len; int main(){ cin>>n; a[++len]=1; for(int i=1;i<=n;i++){ for(int j=1;j<=len;j++){ a[j]=a[j]*i+x; x=a[j]/10; a[j]%=10; } while(x){ a[++len]=x%10; x/=10; } } for(int i=len;i>=1;i--){ cout<<a[i]; } }
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0
#include<bits/stdc++.h> using namespace std; const int N=1e6+10; long long a,b,c,d,e,f,g; double x,y,z; int n[N]; int main(){ cin>>a; n[++b]=1; for(int i=1;i<=a;i++){ for(int j=1;j<=b;j++){ n[j]=n[j]*i+c; c=n[j]/10; n[j]%=10; } while(c){ n[++b]=c%10; c/=10; } } for(int i=b;i>=1;i--)cout<<n[i]; return 0; }
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信息
- ID
- 1198
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- 6
- 标签
- 递交数
- 222
- 已通过
- 67
- 上传者