#include <bits/stdc++.h> using namespace std; int main(){ int a,b,sum = 0; cin >> a >> b; for(int i = a;i <= b;i++){ int t = i; while(t > 0){ sum += t % 10; t /= 10; } } cout << sum; }

#include <iostream> using namespace std; int main() { int a,b,sum=0; cin>>a>>b; for(int i=a;i<=b;i++){ int t=i; while(t>0){ sum+=t%10; t/=10; } } cout<<sum<<endl; return 0; }

0<=a<=b<=1000 这题还算简单，新手自我修养的第二天，话不多说，上代码

附注释题解： 求个赞qwq

对这个区间的每一个数字分解每一位数位，并加入sumn中，最后输出。 复杂度： 设∣n∣|n|∣n∣为该数字数位：O((a−b)∣b∣)O((a-b)|b|)O((a−b)∣b∣)

正经题解

#include <iostream> #include <cstdio> #include<string> using namespace std; int main() { int a,b,c,d=0; cin>>a>>b; for(int i=a;i<=b;i++){ a=i; while(a>0){ c=a; c%=10; d+=c; a/=10; } } cout<<d; return 0; }

#include <iostream> using namespace std; int main() { int a, b; cin >> a >> b; // 读取输入的两个整数 }

#include <bits/stdc++.h> using namespace std; int main(){ int a,b,sum = 0; cin >> a >> b; for(int i = a;i <= b;i++){ int t = i; while(t > 0){ sum += t % 10; t /= 10; } } cout << sum; }

#include<iostream> using namespace std; int a,b,sum,t; int main(){ cin>>a>>b; for(int i=a;i<=b;i++){ t=i; while(t){ sum+=t%10; t/=10; } } cout<<sum; }

#include <iostream> using namespace std; int main(){ int a,b,sum = 0; cin>>a>>b; for(int i=a;i <=b;i++){ int t=i; while(t>0){ sum += t%10; t/=10; } } cout <<sum; return 0; }

#include <bits/stdc++.h> using namespace std; int main(){ int a,b,sum = 0; cin >> a >> b; for(int i = a;i <= b;i++){ int t = i; while(t > 0){ sum += t % 10; t /= 10; } } cout << sum; }

#include <bits/stdc++.h> using namespace std; int main(){ int a,b,c=0; cin>>a>>b; for(int i=a;i<=b;i++){ int t=i; while(t){ c+=t%10; t/=10; } } cout<<c; return 0; }

//#include<iostream> #include<cstdio> //#include<algorithm> //#include<cmath> //#include<string.h> using namespace std; int main() { int a,b,sum=0; scanf("%d%d",&a,&b); for(int i=a;i<=b;i++){ int c=i; while(c){ sum+=c%10; c/=10; } } printf("%d",sum); return 0; }