#include<bits/stdc++.h> using namespace std; int main() { long s, a, b, c; cin >> s; a = s / 1000000; b = s / 1000 % 1000; c = s % 1000; cout << abc; return 0; }

#include<bits/stdc++.h> int a,b,c,n,d;int a_(int a1,int b1,int c1,int n1){d=1000;a=n1/d/d;b=n1/d%d;c=n1%d;return abc;}using namespace std;int main(){cin>>n;cout<<a_(0,0,0,n);}

#include <bits/stdc++.h> using namespace std; int main() { long long a,b,c,sum; cin>>sum; a=sum%1000; sum/=1000; b=sum%1000; sum/=1000; c=sum%1000; sum/=1000; cout<<abc; return 0; }

小码君串台次数我是数过的

感觉大家都不会格式化输入 #include<iostream> using namespace std; int main(){ int a,b,c; scanf("%3d%3d%3d",&a,&b,&c); cout<<abc; return 0; }

正确答案如下

#include <iostream> using namespace std; int main(){ long long a; cin>>a; int x=a%1000; int y=(a%1000000)/1000; int z=a/1000000; cout<<xyz; return 0; }

#include<bits/stdc++.h> using namespace std; int main(){ int a,b,c,d; cin>>d; a=d/1000000%1000; b=d/1000%1000; c=d%1000; cout<<abc; }

#include <bits/stdc++.h> using namespace std; int main(){ long n,a,b,c; cin>>n; a=n%1000; b=n/1000%1000; c=n/1000000; cout<<abc; return 0; }

#include<iostream> using namespace std; int main(){ long long number,a,b,c,g; cin>>number; a=number%1000; b=number/1000%1000; c=number/1000000; cout<<abc; return 0; }

#include<bits/stdc++.h> int a,b,c,n,d;int a_(int a1,int b1,int c1,int n1){d=1000;a=n1/d/d;b=n1/d%d;c=n1%d;return abc;}using namespace std;int main(){cin>>n;cout<<a_(0,0,0,n);}

#include <iostream> using namespace std;int main(){;int a;cin>>a;cout<<(a%1000)*int(((a%1000000-a%1000)*0.001))*int(((a%1000000000-a%1000000)*0.000001));return 0;}

挺烂的 number = input() a = int(number[0]) * 100 + int(number[1]) * 10 + int(number[2]) b = int(number[3]) * 100 + int(number[4]) * 10 + int(number[5]) c = int(number[6]) * 100 + int(number[7]) * 10 + int(number[8]) print(abc)

上题解 #include<iostream> using namespace std; int main(){ long long number,a,b,c,g; cin>>number; a=number%1000; b=number/1000%1000; c=number/1000000; cout<<abc; return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int n; cin >> n; int h3 = n % 1000; n = n / 1000; int z3 = n % 1000; n = n / 1000; int q3 = n % 1000; cout << h3z3q3; return 0; }