#include<bits/stdc++.h>//最优解 using namespace std; double factorial(int n) { double fact = 1; for (int i = 1; i <= n; ++i) fact *= i; return fact; } int main() { int n; double f = 1; cin >> n; for (int i = 1; i <= n; ++i) { double inv_factorial = 1.0 / factorial(i); if (i % 2 == 1) f -= inv_factorial; else f += inv_factorial; } cout << fixed << setprecision(8) << f << endl; return 0; }

ac wa 9999999999999999999999999999999999999999999999999999999999999999999 0 #include<iostream> #include<cstdio> using namespace std; int main(){ int n; double f=1; cin >> n; for(int i=1;i<=n;i++){ int x=1; for(int j=1;j<=i;j++){ x*=j; } if(i%21) f-=1.0/x; else if(i%20) f+=1.0/x; }

#include <iostream> using namespace std; int main(){ int n; cin>>n; double a=1,b=1,c=1; for(int i=1;i<=n;i++){ a=a+1/b; b=b*(i+1); } for(int i=1;i<=n;i=i+2){ if(i>1) c=c*(i-1)*i; a=a-2/c; }

#include <bits/stdc++.h> using namespace std; int main(){ double n; cin >> n; if(n1){cout << "0.00000000";} else if(n2){cout << "0.50000000";} else if(n3){cout << "0.33333333";} else if(n4){cout << "0.37500000";} else if(n5){cout << "0.36666667";} else if(n6){cout << "0.36805556";} else if(n7){cout << "0.36785714";} else if(n8){cout << "0.36788194";} else if(n9){cout << "0.36787919";} else if(n10){cout << "0.36787946";} else if(n11){cout << "0.36787944";} else if(n12){cout << "0.36787944";} else {cout << "0.36787944";} return 0; }

这道题自认为最简单的方法，是个人学了for循环都看得懂的，所以还不敢快点赞！(小心半夜我把你家马桶盖偷了)

#include<bits/stdc++.h> using namespace std; int n,f=1; double x=2.0,cur=1.0; double ni; int main() { cin>>n; ni=0.0; for(int i=2;i<=n;i++) { cur=cur*x; x+=1.0; if(f==1) { ni+=1.0/cur; f=0; } else { ni-=1.0/cur; f=1; } } cout<<fixed<<setprecision(8)<<ni<<endl; return 0; } 我自己都看不懂我在写什么

#include <bits/stdc++.h> using namespace std; int jc(int x){ if (x==1) return 1; return x*jc(x-1); } int main(){ }

#include<iostream> #include<cstdio> #include<math.h> using namespace std; int main(){ int n, y; double f = 1; cin>>n; for(int i=1;i<=n;i++){ y = 1; for(int x=1;x<=i;x++){ y *= x; } if(i%2 == 0){ f += pow(y, -1); } else if(i%2 == 1){ f -= pow(y, -1); } } printf("%.8f", f); return 0; }

#include <bits/stdc++.h> using namespace std; int calculate(int n) { long long sum = 1; for (int i = n; i >= 1; i--) { sum *= i; } return sum; } int main() { int n; cin >> n; double f = 1; if (n == 1) { cout << "0.00000000"; return 0; } for (int i = 1; i <= n; i++) { if (i % 2 == 0) { f = f + (1.0 / calculate(i)); } else { f = f - (1.0 / calculate(i)); } } cout << fixed << setprecision(8) << f; }

#include <bits/stdc++.h> using namespace std; int jc(int x){ if (x==1) return 1; return x*jc(x-1); } int main(){ int n; double f=0; cin>>n; for (int i=1; i<=n; i++){ if (i%2!=0){ f -= 1.01/jc(i); } else f += 1.01/jc(i); } printf("%.8f",1+f); return 0; }

15行的代码！！

#include<iostream> #include<cstdio> using namespace std; int main(){ int n; double f=1; cin >> n; for(int i=1;i<=n;i++){ int x=1; for(int j=1;j<=i;j++){ x*=j; } if(i%21) f-=1.0/x; else if(i%20) f+=1.0/x; }

#include <iostream> #include <cstdio> using namespace std; int main(){ double F=1; int n; int i=1,j=1; double jiecheng=1; cin>>n; while(j<=n){ jiecheng=1; while(i<=j){jiecheng*=i;i++;} i=1; if(j%2!=0){F-=(1/jiecheng);}else{F+=(1/jiecheng);} j++; } printf("%.8f",F); }